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How to Solve Any Physics Problem

Last Updated: July 21, 2023 Fact Checked

This article was co-authored by Sean Alexander, MS . Sean Alexander is an Academic Tutor specializing in teaching mathematics and physics. Sean is the Owner of Alexander Tutoring, an academic tutoring business that provides personalized studying sessions focused on mathematics and physics. With over 15 years of experience, Sean has worked as a physics and math instructor and tutor for Stanford University, San Francisco State University, and Stanbridge Academy. He holds a BS in Physics from the University of California, Santa Barbara and an MS in Theoretical Physics from San Francisco State University. This article has been fact-checked, ensuring the accuracy of any cited facts and confirming the authority of its sources. This article has been viewed 319,017 times.

Baffled as to where to begin with a physics problem? There is a very simply and logical flow process to solving any physics problem.

Step 1 Calm down.

  • Ask yourself if your answers make sense. If the numbers look absurd (for example, you get that a rock dropped off a 50-meter cliff moves with the speed of only 0.00965 meters per second when it hits the ground), you made a mistake somewhere.
  • Don't forget to include the units into your answers, and always keep track of them. So, if you are solving for velocity and get your answer in seconds, that is a sign that something went wrong, because it should be in meters per second.
  • Plug your answers back into the original equations to make sure you get the same number on both sides.

Step 10 Put a box, circle, or underline your answer to make your work neat.

Community Q&A

Community Answer

Video . By using this service, some information may be shared with YouTube.

  • Many people report that if they leave a problem for a while and come back to it later, they find they have a new perspective on it and can sometimes see an easy way to the answer that they did not notice before. Thanks Helpful 249 Not Helpful 47
  • Try to understand the problem first. Thanks Helpful 186 Not Helpful 51
  • Remember, the physics part of the problem is figuring out what you are solving for, drawing the diagram, and remembering the formulae. The rest is just use of algebra, trigonometry, and/or calculus, depending on the difficulty of your course. Thanks Helpful 114 Not Helpful 34

how to solve a physics problem

  • Physics is not easy to grasp for many people, so do not get bent out of shape over a problem. Thanks Helpful 100 Not Helpful 24
  • If an instructor tells you to draw a free body diagram, be sure that that is exactly what you draw. Thanks Helpful 88 Not Helpful 24

Things You'll Need

  • A Writing Utensil (preferably a pencil or erasable pen of sorts)
  • Calculator with all the functions you need for your exam
  • An understanding of the equations needed to solve the problems. Or a list of them will suffice if you are just trying to get through the course alive.

You Might Also Like

Convert Kelvin to Fahrenheit or Celsius

Expert Interview

how to solve a physics problem

Thanks for reading our article! If you’d like to learn more about teaching, check out our in-depth interview with Sean Alexander, MS .

  • ↑ https://iopscience.iop.org/article/10.1088/1361-6404/aa9038
  • ↑ https://physics.wvu.edu/files/d/ce78505d-1426-4d68-8bb2-128d8aac6b1b/expertapproachtosolvingphysicsproblems.pdf
  • ↑ https://www.brighthubeducation.com/science-homework-help/42596-tips-to-choosing-the-correct-physics-formula/

About This Article

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  • 6.1 Solving Problems with Newton’s Laws
  • Introduction
  • 1.1 The Scope and Scale of Physics
  • 1.2 Units and Standards
  • 1.3 Unit Conversion
  • 1.4 Dimensional Analysis
  • 1.5 Estimates and Fermi Calculations
  • 1.6 Significant Figures
  • 1.7 Solving Problems in Physics
  • Key Equations
  • Conceptual Questions
  • Additional Problems
  • Challenge Problems
  • 2.1 Scalars and Vectors
  • 2.2 Coordinate Systems and Components of a Vector
  • 2.3 Algebra of Vectors
  • 2.4 Products of Vectors
  • 3.1 Position, Displacement, and Average Velocity
  • 3.2 Instantaneous Velocity and Speed
  • 3.3 Average and Instantaneous Acceleration
  • 3.4 Motion with Constant Acceleration
  • 3.5 Free Fall
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.3 Projectile Motion
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.1 Conditions for Static Equilibrium
  • 12.2 Examples of Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.6 Bernoulli’s Equation
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
  • 16.2 Mathematics of Waves
  • 16.3 Wave Speed on a Stretched String
  • 16.4 Energy and Power of a Wave
  • 16.5 Interference of Waves
  • 16.6 Standing Waves and Resonance
  • 17.1 Sound Waves
  • 17.2 Speed of Sound
  • 17.3 Sound Intensity
  • 17.4 Normal Modes of a Standing Sound Wave
  • 17.5 Sources of Musical Sound
  • 17.7 The Doppler Effect
  • 17.8 Shock Waves
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

Learning Objectives

By the end of this section, you will be able to:

  • Apply problem-solving techniques to solve for quantities in more complex systems of forces
  • Use concepts from kinematics to solve problems using Newton’s laws of motion
  • Solve more complex equilibrium problems
  • Solve more complex acceleration problems
  • Apply calculus to more advanced dynamics problems

Success in problem solving is necessary to understand and apply physical principles. We developed a pattern of analyzing and setting up the solutions to problems involving Newton’s laws in Newton’s Laws of Motion ; in this chapter, we continue to discuss these strategies and apply a step-by-step process.

Problem-Solving Strategies

We follow here the basics of problem solving presented earlier in this text, but we emphasize specific strategies that are useful in applying Newton’s laws of motion . Once you identify the physical principles involved in the problem and determine that they include Newton’s laws of motion, you can apply these steps to find a solution. These techniques also reinforce concepts that are useful in many other areas of physics. Many problem-solving strategies are stated outright in the worked examples, so the following techniques should reinforce skills you have already begun to develop.

Problem-Solving Strategy

Applying newton’s laws of motion.

  • Identify the physical principles involved by listing the givens and the quantities to be calculated.
  • Sketch the situation, using arrows to represent all forces.
  • Determine the system of interest. The result is a free-body diagram that is essential to solving the problem.
  • Apply Newton’s second law to solve the problem. If necessary, apply appropriate kinematic equations from the chapter on motion along a straight line.
  • Check the solution to see whether it is reasonable.

Let’s apply this problem-solving strategy to the challenge of lifting a grand piano into a second-story apartment. Once we have determined that Newton’s laws of motion are involved (if the problem involves forces), it is particularly important to draw a careful sketch of the situation. Such a sketch is shown in Figure 6.2 (a). Then, as in Figure 6.2 (b), we can represent all forces with arrows. Whenever sufficient information exists, it is best to label these arrows carefully and make the length and direction of each correspond to the represented force.

As with most problems, we next need to identify what needs to be determined and what is known or can be inferred from the problem as stated, that is, make a list of knowns and unknowns. It is particularly crucial to identify the system of interest, since Newton’s second law involves only external forces. We can then determine which forces are external and which are internal, a necessary step to employ Newton’s second law. (See Figure 6.2 (c).) Newton’s third law may be used to identify whether forces are exerted between components of a system (internal) or between the system and something outside (external). As illustrated in Newton’s Laws of Motion , the system of interest depends on the question we need to answer. Only forces are shown in free-body diagrams, not acceleration or velocity. We have drawn several free-body diagrams in previous worked examples. Figure 6.2 (c) shows a free-body diagram for the system of interest. Note that no internal forces are shown in a free-body diagram.

Once a free-body diagram is drawn, we apply Newton’s second law. This is done in Figure 6.2 (d) for a particular situation. In general, once external forces are clearly identified in free-body diagrams, it should be a straightforward task to put them into equation form and solve for the unknown, as done in all previous examples. If the problem is one-dimensional—that is, if all forces are parallel—then the forces can be handled algebraically. If the problem is two-dimensional, then it must be broken down into a pair of one-dimensional problems. We do this by projecting the force vectors onto a set of axes chosen for convenience. As seen in previous examples, the choice of axes can simplify the problem. For example, when an incline is involved, a set of axes with one axis parallel to the incline and one perpendicular to it is most convenient. It is almost always convenient to make one axis parallel to the direction of motion, if this is known. Generally, just write Newton’s second law in components along the different directions. Then, you have the following equations:

(If, for example, the system is accelerating horizontally, then you can then set a y = 0 . a y = 0 . ) We need this information to determine unknown forces acting on a system.

As always, we must check the solution. In some cases, it is easy to tell whether the solution is reasonable. For example, it is reasonable to find that friction causes an object to slide down an incline more slowly than when no friction exists. In practice, intuition develops gradually through problem solving; with experience, it becomes progressively easier to judge whether an answer is reasonable. Another way to check a solution is to check the units. If we are solving for force and end up with units of millimeters per second, then we have made a mistake.

There are many interesting applications of Newton’s laws of motion, a few more of which are presented in this section. These serve also to illustrate some further subtleties of physics and to help build problem-solving skills. We look first at problems involving particle equilibrium, which make use of Newton’s first law, and then consider particle acceleration, which involves Newton’s second law.

Particle Equilibrium

Recall that a particle in equilibrium is one for which the external forces are balanced. Static equilibrium involves objects at rest, and dynamic equilibrium involves objects in motion without acceleration, but it is important to remember that these conditions are relative. For example, an object may be at rest when viewed from our frame of reference, but the same object would appear to be in motion when viewed by someone moving at a constant velocity. We now make use of the knowledge attained in Newton’s Laws of Motion , regarding the different types of forces and the use of free-body diagrams, to solve additional problems in particle equilibrium .

Example 6.1

Different tensions at different angles.

Thus, as you might expect,

This gives us the following relationship:

Note that T 1 T 1 and T 2 T 2 are not equal in this case because the angles on either side are not equal. It is reasonable that T 2 T 2 ends up being greater than T 1 T 1 because it is exerted more vertically than T 1 . T 1 .

Now consider the force components along the vertical or y -axis:

This implies

Substituting the expressions for the vertical components gives

There are two unknowns in this equation, but substituting the expression for T 2 T 2 in terms of T 1 T 1 reduces this to one equation with one unknown:

which yields

Solving this last equation gives the magnitude of T 1 T 1 to be

Finally, we find the magnitude of T 2 T 2 by using the relationship between them, T 2 = 1.225 T 1 T 2 = 1.225 T 1 , found above. Thus we obtain


Particle acceleration.

We have given a variety of examples of particles in equilibrium. We now turn our attention to particle acceleration problems, which are the result of a nonzero net force. Refer again to the steps given at the beginning of this section, and notice how they are applied to the following examples.

Example 6.2

Drag force on a barge.

The drag of the water F → D F → D is in the direction opposite to the direction of motion of the boat; this force thus works against F → app , F → app , as shown in the free-body diagram in Figure 6.4 (b). The system of interest here is the barge, since the forces on it are given as well as its acceleration. Because the applied forces are perpendicular, the x - and y -axes are in the same direction as F → 1 F → 1 and F → 2 . F → 2 . The problem quickly becomes a one-dimensional problem along the direction of F → app F → app , since friction is in the direction opposite to F → app . F → app . Our strategy is to find the magnitude and direction of the net applied force F → app F → app and then apply Newton’s second law to solve for the drag force F → D . F → D .

The angle is given by

From Newton’s first law, we know this is the same direction as the acceleration. We also know that F → D F → D is in the opposite direction of F → app , F → app , since it acts to slow down the acceleration. Therefore, the net external force is in the same direction as F → app , F → app , but its magnitude is slightly less than F → app . F → app . The problem is now one-dimensional. From the free-body diagram, we can see that

However, Newton’s second law states that

This can be solved for the magnitude of the drag force of the water F D F D in terms of known quantities:

Substituting known values gives

The direction of F → D F → D has already been determined to be in the direction opposite to F → app , F → app , or at an angle of 53 ° 53 ° south of west.

In Newton’s Laws of Motion , we discussed the normal force , which is a contact force that acts normal to the surface so that an object does not have an acceleration perpendicular to the surface. The bathroom scale is an excellent example of a normal force acting on a body. It provides a quantitative reading of how much it must push upward to support the weight of an object. But can you predict what you would see on the dial of a bathroom scale if you stood on it during an elevator ride? Will you see a value greater than your weight when the elevator starts up? What about when the elevator moves upward at a constant speed? Take a guess before reading the next example.

Example 6.3

What does the bathroom scale read in an elevator.

From the free-body diagram, we see that F → net = F → s − w → , F → net = F → s − w → , so we have

Solving for F s F s gives us an equation with only one unknown:

or, because w = m g , w = m g , simply

No assumptions were made about the acceleration, so this solution should be valid for a variety of accelerations in addition to those in this situation. ( Note: We are considering the case when the elevator is accelerating upward. If the elevator is accelerating downward, Newton’s second law becomes F s − w = − m a . F s − w = − m a . )

  • We have a = 1.20 m/s 2 , a = 1.20 m/s 2 , so that F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) F s = ( 75.0 kg ) ( 9.80 m/s 2 ) + ( 75.0 kg ) ( 1.20 m/s 2 ) yielding F s = 825 N . F s = 825 N .
  • Now, what happens when the elevator reaches a constant upward velocity? Will the scale still read more than his weight? For any constant velocity—up, down, or stationary—acceleration is zero because a = Δ v Δ t a = Δ v Δ t and Δ v = 0 . Δ v = 0 . Thus, F s = m a + m g = 0 + m g F s = m a + m g = 0 + m g or F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , F s = ( 75.0 kg ) ( 9.80 m/s 2 ) , which gives F s = 735 N . F s = 735 N .

Thus, the scale reading in the elevator is greater than his 735-N (165-lb.) weight. This means that the scale is pushing up on the person with a force greater than his weight, as it must in order to accelerate him upward. Clearly, the greater the acceleration of the elevator, the greater the scale reading, consistent with what you feel in rapidly accelerating versus slowly accelerating elevators. In Figure 6.5 (b), the scale reading is 735 N, which equals the person’s weight. This is the case whenever the elevator has a constant velocity—moving up, moving down, or stationary.

Check Your Understanding 6.1

Now calculate the scale reading when the elevator accelerates downward at a rate of 1.20 m/s 2 . 1.20 m/s 2 .

The solution to the previous example also applies to an elevator accelerating downward, as mentioned. When an elevator accelerates downward, a is negative, and the scale reading is less than the weight of the person. If a constant downward velocity is reached, the scale reading again becomes equal to the person’s weight. If the elevator is in free fall and accelerating downward at g , then the scale reading is zero and the person appears to be weightless.

Example 6.4

Two attached blocks.

For block 1: T → + w → 1 + N → = m 1 a → 1 T → + w → 1 + N → = m 1 a → 1

For block 2: T → + w → 2 = m 2 a → 2 . T → + w → 2 = m 2 a → 2 .

Notice that T → T → is the same for both blocks. Since the string and the pulley have negligible mass, and since there is no friction in the pulley, the tension is the same throughout the string. We can now write component equations for each block. All forces are either horizontal or vertical, so we can use the same horizontal/vertical coordinate system for both objects

When block 1 moves to the right, block 2 travels an equal distance downward; thus, a 1 x = − a 2 y . a 1 x = − a 2 y . Writing the common acceleration of the blocks as a = a 1 x = − a 2 y , a = a 1 x = − a 2 y , we now have

From these two equations, we can express a and T in terms of the masses m 1 and m 2 , and g : m 1 and m 2 , and g :

Check Your Understanding 6.2

Calculate the acceleration of the system, and the tension in the string, when the masses are m 1 = 5.00 kg m 1 = 5.00 kg and m 2 = 3.00 kg . m 2 = 3.00 kg .

Example 6.5

Atwood machine.

  • We have For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . For m 1 , ∑ F y = T − m 1 g = m 1 a . For m 2 , ∑ F y = T − m 2 g = − m 2 a . (The negative sign in front of m 2 a m 2 a indicates that m 2 m 2 accelerates downward; both blocks accelerate at the same rate, but in opposite directions.) Solve the two equations simultaneously (subtract them) and the result is ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . ( m 2 − m 1 ) g = ( m 1 + m 2 ) a . Solving for a : a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 . a = m 2 − m 1 m 1 + m 2 g = 4 kg − 2 kg 4 kg + 2 kg ( 9.8 m/s 2 ) = 3.27 m/s 2 .
  • Observing the first block, we see that T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N . T − m 1 g = m 1 a T = m 1 ( g + a ) = ( 2 kg ) ( 9.8 m/s 2 + 3.27 m/s 2 ) = 26.1 N .

Check Your Understanding 6.3

Determine a general formula in terms of m 1 , m 2 m 1 , m 2 and g for calculating the tension in the string for the Atwood machine shown above.

Newton’s Laws of Motion and Kinematics

Physics is most interesting and most powerful when applied to general situations that involve more than a narrow set of physical principles. Newton’s laws of motion can also be integrated with other concepts that have been discussed previously in this text to solve problems of motion. For example, forces produce accelerations, a topic of kinematics , and hence the relevance of earlier chapters.

When approaching problems that involve various types of forces, acceleration, velocity, and/or position, listing the givens and the quantities to be calculated will allow you to identify the principles involved. Then, you can refer to the chapters that deal with a particular topic and solve the problem using strategies outlined in the text. The following worked example illustrates how the problem-solving strategy given earlier in this chapter, as well as strategies presented in other chapters, is applied to an integrated concept problem.

Example 6.6

What force must a soccer player exert to reach top speed.

  • We are given the initial and final velocities (zero and 8.00 m/s forward); thus, the change in velocity is Δ v = 8.00 m/s Δ v = 8.00 m/s . We are given the elapsed time, so Δ t = 2.50 s . Δ t = 2.50 s . The unknown is acceleration, which can be found from its definition: a = Δ v Δ t . a = Δ v Δ t . Substituting the known values yields a = 8.00 m/s 2.50 s = 3.20 m/s 2 . a = 8.00 m/s 2.50 s = 3.20 m/s 2 .
  • Here we are asked to find the average force the ground exerts on the runner to produce this acceleration. (Remember that we are dealing with the force or forces acting on the object of interest.) This is the reaction force to that exerted by the player backward against the ground, by Newton’s third law. Neglecting air resistance, this would be equal in magnitude to the net external force on the player, since this force causes her acceleration. Since we now know the player’s acceleration and are given her mass, we can use Newton’s second law to find the force exerted. That is, F net = m a . F net = m a . Substituting the known values of m and a gives F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N . F net = ( 70.0 kg ) ( 3.20 m/s 2 ) = 224 N .

This is a reasonable result: The acceleration is attainable for an athlete in good condition. The force is about 50 pounds, a reasonable average force.

Check Your Understanding 6.4

The soccer player stops after completing the play described above, but now notices that the ball is in position to be stolen. If she now experiences a force of 126 N to attempt to steal the ball, which is 2.00 m away from her, how long will it take her to get to the ball?

Example 6.7

What force acts on a model helicopter.

The magnitude of the force is now easily found:

Check Your Understanding 6.5

Find the direction of the resultant for the 1.50-kg model helicopter.

Example 6.8

Baggage tractor.

  • ∑ F x = m system a x ∑ F x = m system a x and ∑ F x = 820.0 t , ∑ F x = 820.0 t , so 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . 820.0 t = ( 650.0 + 250.0 + 150.0 ) a a = 0.7809 t . Since acceleration is a function of time, we can determine the velocity of the tractor by using a = d v d t a = d v d t with the initial condition that v 0 = 0 v 0 = 0 at t = 0 . t = 0 . We integrate from t = 0 t = 0 to t = 3 : t = 3 : d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s . d v = a d t , ∫ 0 3 d v = ∫ 0 3.00 a d t = ∫ 0 3.00 0.7809 t d t , v = 0.3905 t 2 ] 0 3.00 = 3.51 m/s .
  • Refer to the free-body diagram in Figure 6.8 (b). ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N . ∑ F x = m tractor a x 820.0 t − T = m tractor ( 0.7805 ) t ( 820.0 ) ( 3.00 ) − T = ( 650.0 ) ( 0.7805 ) ( 3.00 ) T = 938 N .

Recall that v = d s d t v = d s d t and a = d v d t a = d v d t . If acceleration is a function of time, we can use the calculus forms developed in Motion Along a Straight Line , as shown in this example. However, sometimes acceleration is a function of displacement. In this case, we can derive an important result from these calculus relations. Solving for dt in each, we have d t = d s v d t = d s v and d t = d v a . d t = d v a . Now, equating these expressions, we have d s v = d v a . d s v = d v a . We can rearrange this to obtain a d s = v d v . a d s = v d v .

Example 6.9

Motion of a projectile fired vertically.

The acceleration depends on v and is therefore variable. Since a = f ( v ) , a = f ( v ) , we can relate a to v using the rearrangement described above,

We replace ds with dy because we are dealing with the vertical direction,

We now separate the variables ( v ’s and dv ’s on one side; dy on the other):

Thus, h = 114 m . h = 114 m .

Check Your Understanding 6.6

If atmospheric resistance is neglected, find the maximum height for the mortar shell. Is calculus required for this solution?


Explore the forces at work in this simulation when you try to push a filing cabinet. Create an applied force and see the resulting frictional force and total force acting on the cabinet. Charts show the forces, position, velocity, and acceleration vs. time. View a free-body diagram of all the forces (including gravitational and normal forces).

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Example Physics Problems and Solutions

Equilibrium Example Problem 1

Learning how to solve physics problems is a big part of learning physics. Here’s a collection of example physics problems and solutions to help you tackle problems sets and understand concepts and how to work with formulas:

Physics Homework Tips Physics homework can be challenging! Get tips to help make the task a little easier.

Unit Conversion Examples

There are now too many unit conversion examples to list in this space. This Unit Conversion Examples page is a more comprehensive list of worked example problems.

Newton’s Equations of Motion Example Problems

Equations of Motion – Constant Acceleration Example This equations of motion example problem consist of a sliding block under constant acceleration. It uses the equations of motion to calculate the position and velocity of a given time and the time and position of a given velocity.

Equations of Motion Example Problem – Constant Acceleration This example problem uses the equations of motion for constant acceleration to find the position, velocity, and acceleration of a breaking vehicle.

Equations of Motion Example Problem – Interception

This example problem uses the equations of motion for constant acceleration to calculate the time needed for one vehicle to intercept another vehicle moving at a constant velocity.

well drop setup illustration

Vertical Motion Example Problem – Coin Toss Here’s an example applying the equations of motion under constant acceleration to determine the maximum height, velocity and time of flight for a coin flipped into a well. This problem could be modified to solve any object tossed vertically or dropped off a tall building or any height. This type of problem is a common equation of motion homework problem.

Projectile Motion Example Problem This example problem shows how to find different variables associated with parabolic projectile motion.


Accelerometer and Inertia Example Problem Accelerometers are devices to measure or detect acceleration by measuring the changes that occur as a system experiences an acceleration. This example problem uses one of the simplest forms of an accelerometer, a weight hanging from a stiff rod or wire. As the system accelerates, the hanging weight is deflected from its rest position. This example derives the relationship between that angle, the acceleration and the acceleration due to gravity. It then calculates the acceleration due to gravity of an unknown planet.

Weight In An Elevator Have you ever wondered why you feel slightly heavier in an elevator when it begins to move up? Or why you feel lighter when the elevator begins to move down? This example problem explains how to find your weight in an accelerating elevator and how to find the acceleration of an elevator using your weight on a scale.

Equilibrium Example Problem This example problem shows how to determine the different forces in a system at equilibrium. The system is a block suspended from a rope attached to two other ropes.

Equilibrium Cat 1

Equilibrium Example Problem – Balance This example problem highlights the basics of finding the forces acting on a system in mechanical equilibrium.

Force of Gravity Example This physics problem and solution shows how to apply Newton’s equation to calculate the gravitational force between the Earth and the Moon.

Coupled Systems Example Problems

Atwood Machine

Coupled systems are two or more separate systems connected together. The best way to solve these types of problems is to treat each system separately and then find common variables between them. Atwood Machine The Atwood Machine is a coupled system of two weights sharing a connecting string over a pulley. This example problem shows how to find the acceleration of an Atwood system and the tension in the connecting string. Coupled Blocks – Inertia Example This example problem is similar to the Atwood machine except one block is resting on a frictionless surface perpendicular to the other block. This block is hanging over the edge and pulling down on the coupled string. The problem shows how to calculate the acceleration of the blocks and the tension in the connecting string.

Friction Example Problems

friction slide setup

These example physics problems explain how to calculate the different coefficients of friction.

Friction Example Problem – Block Resting on a Surface Friction Example Problem – Coefficient of Static Friction Friction Example Problem – Coefficient of Kinetic Friction Friction and Inertia Example Problem

Momentum and Collisions Example Problems

Desktop Momentum Balls Toy

These example problems show how to calculate the momentum of moving masses.

Momentum and Impulse Example Finds the momentum before and after a force acts on a body and determine the impulse of the force.

Elastic Collision Example Shows how to find the velocities of two masses after an elastic collision.

It Can Be Shown – Elastic Collision Math Steps Shows the math to find the equations expressing the final velocities of two masses in terms of their initial velocities.

Simple Pendulum Example Problems

how to solve a physics problem

These example problems show how to use the period of a pendulum to find related information.

Find the Period of a Simple Pendulum Find the period if you know the length of a pendulum and the acceleration due to gravity.

Find the Length of a Simple Pendulum Find the length of the pendulum when the period and acceleration due to gravity is known.

Find the Acceleration due to Gravity Using A Pendulum Find ‘g’ on different planets by timing the period of a known pendulum length.

Harmonic Motion and Waves Example Problems

Hooke's Law Forces

These example problems all involve simple harmonic motion and wave mechanics.

Energy and Wavelength Example This example shows how to determine the energy of a photon of a known wavelength.

Hooke’s Law Example Problem An example problem involving the restoring force of a spring.

Wavelength and Frequency Calculations See how to calculate wavelength if you know frequency and vice versa, for light, sound, or other waves.

Heat and Energy Example Problems

Heat of Fusion Example Problem Two example problems using the heat of fusion to calculate the energy required for a phase change.

Specific Heat Example Problem This is actually 3 similar example problems using the specific heat equation to calculate heat, specific heat, and temperature of a system.

Heat of Vaporization Example Problems Two example problems using or finding the heat of vaporization.

Ice to Steam Example Problem Classic problem melting cold ice to make hot steam. This problem brings all three of the previous example problems into one problem to calculate heat changes over phase changes.

Charge and Coulomb Force Example Problems

Setup diagram of Coulomb's Law Example Problem.

Electrical charges generate a coulomb force between themselves proportional to the magnitude of the charges and inversely proportional to the distance between them. Coulomb’s Law Example This example problem shows how to use Coulomb’s Law equation to find the charges necessary to produce a known repulsive force over a set distance. Coulomb Force Example This Coulomb force example shows how to find the number of electrons transferred between two bodies to generate a set amount of force over a short distance.

How To Solve Physics Problems

4 tricks for solving any physics problem

Physics can be intimidating—all those pulleys and protons and projectile motion. If you approach it with the right mindset, however, even the hardest problems are usually easier than you think. When you come up against a tough question, don’t panic. Instead, start with these short, easy tricks to help you work through the problem.  

4 tricks for solving any physics problem:

1. what is the subject.

Just about every physics question is testing specific knowledge. When you read the question ask yourself, is it exploring electricity? Torque? Parabolic motion? Each topic is associated with specific equations and approaches, so recognizing the subject will focus your effort in the right direction. Look for keywords and phrases that reveal the topic. 

2. What are you trying to find?

This simple step can save a lot of time. Before starting to solve the problem, think about what the answer will look like. What are the units; is the final answer going to be in kilograms or liters? Also, consider what other physical quantities might relate to your answer. If you’re trying to find speed, it might be useful to find acceleration, then solve that for speed. Determining restrictions on the answer early also ensures you answer the specific question; a common mistake in physics is solving for the wrong thing. 

3. What do you know?

Think about what details the problem mentions. Unless the question is really bad, they probably gave you exactly the information you need to solve the problem. Don’t be surprised if sometimes this information is coded in language; a problem that mentions a spring with “the mass removed from the end” is telling you something important about the quantities of force. Write down every quantity you know from the problem, then proceed to…

4. What equations can you use?

What equations include the quantities you know and also the one you’re looking for? If you have the mass of an object and a force and you’re trying to find the acceleration, start with F=ma (Newton’s second law). If you’re trying to find the electric field but you have the charge and the distance, try E=q/(4πε*r 2 ). 

If you’re having trouble figuring out which equation to use, go back to our first trick. What equations are associated with the topic? Can you manipulate the quantities you have to fit in any of them? 

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how to solve a physics problem

  • Newton's Laws
  • Einstein's Theory of Special Relativity
  • About Concept Checkers
  • School Pricing
  • Newton's Laws of Motion
  • Newton's First Law
  • Newton's Third Law
  • Horizontally Launched Projectile Problems
  • What is a Projectile?
  • Motion Characteristics of a Projectile
  • Horizontal and Vertical Velocity
  • Horizontal and Vertical Displacement
  • Initial Velocity Components
  • Non-Horizontally Launched Projectile Problems

how to solve a physics problem

There are two basic types of projectile problems that we will discuss in this course. While the general principles are the same for each type of problem, the approach will vary due to the fact the problems differ in terms of their initial conditions. The two types of problems are:

A projectile is launched with an initial horizontal velocity from an elevated position and follows a parabolic path to the ground. Predictable unknowns include the initial speed of the projectile, the initial height of the projectile, the time of flight, and the horizontal distance of the projectile.

Examples of this type of problem are

  • A pool ball leaves a 0.60-meter high table with an initial horizontal velocity of 2.4 m/s. Predict the time required for the pool ball to fall to the ground and the horizontal distance between the table's edge and the ball's landing location.

A soccer ball is kicked horizontally off a 22.0-meter high hill and lands a distance of 35.0 meters from the edge of the hill. Determine the initial horizontal velocity of the soccer ball.

A projectile is launched at an angle to the horizontal and rises upwards to a peak while moving horizontally. Upon reaching the peak, the projectile falls with a motion that is symmetrical to its path upwards to the peak. Predictable unknowns include the time of flight, the horizontal range, and the height of the projectile when it is at its peak.

  • A football is kicked with an initial velocity of 25 m/s at an angle of 45-degrees with the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the football.
  • A long jumper leaves the ground with an initial velocity of 12 m/s at an angle of 28-degrees above the horizontal. Determine the time of flight, the horizontal distance, and the peak height of the long-jumper.

The second problem type will be the subject of the next part of Lesson 2 . In this part of Lesson 2, we will focus on the first type of problem - sometimes referred to as horizontally launched projectile problems. Three common kinematic equations that will be used for both type of problems include the following:

d = v i •t + 0.5*a*t 2 v f = v i + a•t v f 2  = v i 2  + 2*a•d  

Equations for the Horizontal Motion of a Projectile

The above equations work well for motion in one-dimension, but a projectile is usually moving in two dimensions - both horizontally and vertically. Since these two components of motion are independent of each other, two distinctly separate sets of equations are needed - one for the projectile's horizontal motion and one for its vertical motion. Thus, the three equations above are transformed into two sets of three equations. For the horizontal components of motion, the equations are

x = v i x •t + 0.5*a x *t 2

v f x  = v i x  + a x •t

v f x 2  = v i x 2  + 2*a x •x

Of these three equations, the top equation is the most commonly used. An application of projectile concepts to each of these equations would also lead one to conclude that any term with a x in it would cancel out of the equation since a x = 0 m/s/s . Once this cancellation of ax terms is performed, the only equation of usefulness is:

x = v i x •t

Equations for the Vertical Motion of a Projectile

For the vertical components of motion, the three equations are

y = v iy •t + 0.5*a y *t 2

v fy  = v iy  + a y •t

v fy 2  = v iy 2  + 2*a y •y

In each of the above equations, the vertical acceleration of a projectile is known to be -9.8 m/s/s (the acceleration of gravity). Furthermore, for the special case of the first type of problem (horizontally launched projectile problems), v iy = 0 m/s. Thus, any term with v iy in it will cancel out of the equation.

The two sets of three equations above are the kinematic equations that will be used to solve projectile motion problems.

Solving Projectile Problems

To illustrate the usefulness of the above equations in making predictions about the motion of a projectile, consider the solution to the following problem.

The solution of this problem begins by equating the known or given values with the symbols of the kinematic equations - x, y, v ix , v iy , a x , a y , and t. Because horizontal and vertical information is used separately, it is a wise idea to organized the given information in two columns - one column for horizontal information and one column for vertical information. In this case, the following information is either given or implied in the problem statement:

As indicated in the table, the unknown quantity is the horizontal displacement (and the time of flight) of the pool ball. The solution of the problem now requires the selection of an appropriate strategy for using the kinematic equations and the known information to solve for the unknown quantities. It will almost always be the case that such a strategy demands that one of the vertical equations be used to determine the time of flight of the projectile and then one of the horizontal equations be used to find the other unknown quantities (or vice versa - first use the horizontal and then the vertical equation). An organized listing of known quantities (as in the table above) provides cues for the selection of the strategy. For example, the table above reveals that there are three quantities known about the vertical motion of the pool ball. Since each equation has four variables in it, knowledge of three of the variables allows one to calculate a fourth variable. Thus, it would be reasonable that a vertical equation is used with the vertical values to determine time and then the horizontal equations be used to determine the horizontal displacement (x). The first vertical equation (y = v iy •t +0.5•a y •t 2 ) will allow for the determination of the time. Once the appropriate equation has been selected, the physics problem becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

Since the first term on the right side of the equation reduces to 0, the equation can be simplified to

If both sides of the equation are divided by -5.0 m/s/s, the equation becomes

By taking the square root of both sides of the equation, the time of flight can then be determined .

Once the time has been determined, a horizontal equation can be used to determine the horizontal displacement of the pool ball. Recall from the given information , v ix = 2.4 m/s and a x = 0 m/s/s. The first horizontal equation (x = v ix •t + 0.5•a x •t 2 ) can then be used to solve for "x." With the equation selected, the physics problem once more becomes transformed into an algebra problem. By substitution of known values, the equation takes the form of

Since the second term on the right side of the equation reduces to 0, the equation can then be simplified to

The answer to the stated problem is that the pool ball is in the air for 0.35 seconds and lands a horizontal distance of 0.84 m from the edge of the pool table.

The following procedure summarizes the above problem-solving approach.

  • Carefully read the problem and list known and unknown information in terms of the symbols of the kinematic equations. For convenience sake, make a table with horizontal information on one side and vertical information on the other side.
  • Identify the unknown quantity that the problem requests you to solve for.
  • Select either a horizontal or vertical equation to solve for the time of flight of the projectile.
  • With the time determined, use one of the other equations to solve for the unknown. (Usually, if a horizontal equation is used to solve for time, then a vertical equation can be used to solve for the final unknown quantity.)

One caution is in order. The sole reliance upon 4- and 5-step procedures to solve physics problems is always a dangerous approach. Physics problems are usually just that - problems! While problems can often be simplified by the use of short procedures as the one above, not all problems can be solved with the above procedure. While steps 1 and 2 above are critical to your success in solving horizontally launched projectile problems, there will always be a problem that doesn't fit the mold . Problem solving is not like cooking; it is not a mere matter of following a recipe. Rather, problem solving requires careful reading, a firm grasp of conceptual physics, critical thought and analysis, and lots of disciplined practice. Never divorce conceptual understanding and critical thinking from your approach to solving problems.

Check Your Understanding

Use y = v iy • t + 0.5 • a y • t 2 to solve for time; the time of flight is 2.12 seconds.

Now use x = v ix • t + 0.5 • a x • t 2 to solve for v ix

Note that a x is 0 m/s/s so the last term on the right side of the equation cancels. By substituting 35.0 m for x and 2.12 s for t, the v ix can be found to be 16.5 m/s.

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Google wants to solve tricky physics problems with quantum computers

Quantum computers could become more useful now researchers at Google have designed an algorithm that can translate complex physical problems into the language of quantum physics

By Alex Wilkins

19 December 2023

A quantum computer

Quantum computers don’t have many practical uses yet

Bartlomiej Wroblewski/Getty Images

Researchers at Google have created an algorithm that can translate complex physical problems into the language of quantum mechanics, which could make quantum computers able to tackle more tasks.

IBM’s 'Condor' quantum computer has more than 1000 qubits

Once they become powerful enough, quantum computers might become useful for specific jobs, such as breaking encryption or modelling quantum mechanics, but it is still largely unknown how useful they will be for many other scientific problems that classical computers…

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Engineering Mathematics and Sciences

Your virtual study buddy in mathematics, engineering sciences, and civil engineering, problem 6-4: period, angular velocity, and linear velocity of the earth, (a) what is the period of rotation of earth in seconds (b) what is the angular velocity of earth (c) given that earth has a radius of 6.4×10 6 m at its equator, what is the linear velocity at earth’s surface.

The period of a rotating body is the time it takes for 1 full revolution. The Earth rotates about its axis, and complete 1 full revolution in 24 hours. Therefore, the period is

The angular velocity \omega is the rate of change of an angle,

where a rotation \Delta \theta takes place in a time \Delta t .

From the given problem, we are given the following: \Delta \theta = 2\pi \text{radian} = 1 \ \text{revolution} , and \Delta t =24\ \text{hours} = 1440 \ \text{minutes}= 86400 \ \text{seconds} . Therefore, the angular velocity is

We can also express the angular velocity in units of radians per second. That is

The linear velocity v , and the angular velocity \omega are related by the formula

From the given problem, we are given the following values: r=6.4 \times 10^{6} \ \text{meters} , and \omega = 7.27 \times 10^{-5}\ \text{radians/second} . Therefore, the linear velocity at the surface of the earth is

December 22, 2023

The Most Important Unsolved Problem in Computer Science

Here’s a look at the million-dollar math problem at the heart of computation

By Jack Murtagh

mathematic formulas on a computer display, blue text on black screen

alengo/Getty Images

When the Clay Mathematics Institute put individual $1-million prize bounties on seven unsolved mathematical problems , they may have undervalued one entry—by a lot. If mathematicians were to resolve, in the right way, computer science’s “P versus NP” question, the result could be worth worlds more than $1 million—they’d be cracking most online-security systems, revolutionizing science and even mechanistically solving the other six of the so-called Millennium Problems, all of which were chosen in the year 2000. It’s hard to overstate the stakes surrounding the most important unsolved problem in computer science .

P versus NP concerns the apparent asymmetry between finding solutions to problems and verifying solutions to problems. For example, imagine you’re planning a world tour to promote your new book. You pull up Priceline and start testing routes, but each one you try blows your total trip budget. Unfortunately, as the number of cities grows on your worldwide tour, the number of possible routes to check skyrockets exponentially, rapidly making it infeasible even for computers to exhaustively search through every case. But when you complain, your agent writes back with a solution sequence of flights. You can easily verify whether or not their route stays in budget by simply checking that it hits every city and summing the fares to compare against the budget limit. Notice the asymmetry here: finding a solution is hard, but verifying a solution is easy.

The P versus NP question asks whether this asymmetry is real or an illusion. If you can efficiently verify a solution to a problem, does that imply that you can also efficiently find a solution? Perhaps a clever shortcut can circumvent searching through zillions of potential routes. For example, if your agent instead wanted you to find a sequence of flights between two specific remote airports while obeying the budget, you might also throw up your hands at the similarly immense number of possible routes to check, but in fact, this problem contains enough structure that computer scientists have developed a fast procedure (algorithm) for it that bypasses the need for exhaustive search.

You might think this asymmetry is obvious: of course one would sometimes have a harder time finding a solution to a problem than verifying it. But researchers have been surprised before in thinking that that’s the case, only to discover last-minute that the solution is just as easy. So every attempt in which they try to resolve this question for any single scenario only further exposes how monumentally difficult it is to prove one way or another. P versus NP also rears its head everywhere we look in the computational world well beyond the specifics of our travel scenario—so much so that it has come to symbolize a holy grail in our understanding of computation.

In the subfield of theoretical computer science called complexity theory, researchers try to pin down how easily computers can solve various types of problems. P represents the class of problems they can solve efficiently, such as sorting a column of numbers in a spreadsheet or finding the shortest path between two addresses on a map. NP represents the class of problems for which computers can verify solutions efficiently. Our book tour problem, called the Traveling Salesperson Problem by academics, lives in NP because we have an efficient procedure for verifying that our agent’s solution worked.

Notice that NP actually contains P as a subset because solving a problem outright is one way to verify a solution to it. For example, how would you verify that 27 x 89 = 2,403? You would solve the multiplication problem yourself and check that your answer matches the claimed one. We typically depict the relationship between P and NP with a simple Venn diagram:

Venn diagram shows one large circle labeled “NP” encompassing a smaller one labeled “P.” The entire circle is labeled “Problems with solutions that computers can verify easily.” The area inside of P is labeled “Problems with solutions that computers can find easily.” The area in NP but outside of P is labeled “Problems with solutions that computers can verify but not find easily.”

The region inside of NP but not inside of P contains problems that can’t be solved with any known efficient algorithm. (Theoretical computer scientists use a technical definition for “efficient” that can be debated, but it serves as a useful proxy for the colloquial concept.) But we don’t know if that’s because such algorithms don’t exist or we just haven’t mustered the ingenuity to discover them. Here’s another way to phrase the P versus NP question: Are these classes actually distinct? Or does the Venn diagram collapse into one circle? Do all NP problems admit efficient algorithms? Here are some examples of problems in NP that are not currently known to be in P:

  • Given a social network, is there a group of a specified size in which all of the people in it are friends with one another?
  • Given a varied collection of boxes to be shipped, can all of them be fit into a specified number of trucks?
  • Given a sudoku (generalized to n x n puzzle grids), does it have a solution?
  • Given a map, can the countries be colored with only three colors such that no two neighboring countries are the same color?

Ask yourself how you would verify proposed solutions to some of the problems above and then how you would find a solution. Note that approximating a solution or solving a small instance (most of us can solve a 9 x 9 sudoku ) doesn’t suffice. To qualify as solving a problem, an algorithm needs to find an exact solution on all instances, including very large ones.

Each of the problems can be solved via brute-force search (e.g., try every possible coloring of the map and check if any of them work), but the number of cases to try grows exponentially with the size of the problem. This means that if we call the size of the problem n (e.g., the number of countries on the map or the number of boxes to pack into trucks), then the number of cases to check looks something like 2 n . The world’s fastest supercomputers have no hope against exponential growth. Even when n equals 300, a tiny input size by modern data standards, 2 300 exceeds the number of atoms in the observable universe. After hitting “go” on such an algorithm, your computer would display a spinning pinwheel that would outlive you and your descendants.

Thousands of other problems belong on our list. From cell biology to game theory, the P versus NP question reaches into far corners of science and industry. If P = NP (i.e., our Venn diagram dissolves into a single circle) and we obtain fast algorithms for these seemingly hard problems, then the whole digital economy would become vulnerable to collapse. This is because much of the cryptography that secures such things as your credit card number and passwords works by shrouding private information behind computationally difficult problems that can only become easy to solve if you know the secret key. Online security as we know it rests on unproven mathematical assumptions that crumble if P = NP.

Amazingly, we can even cast math itself as an NP problem because we can program computers to efficiently verify proofs. In fact, legendary mathematician Kurt Gödel first posed the P versus NP problem in a letter to his colleague John von Neumann in 1956, and he expressed (in older terminology) that P = NP “would have consequences of the greatest importance. Namely, it would obviously mean that ... the mental work of a mathematician concerning yes-or-no questions could be completely replaced by a machine.”

If you’re a mathematician worried for your job, rest assured that most experts believe that P does not equal NP. Aside from the intuition that sometimes solutions should be harder to find than to verify, thousands of the hardest NP problems that are not known to be in P have sat unsolved across disparate fields, glowing with incentives of fame and fortune, and yet not one person has designed an efficient algorithm for a single one of them.

Of course, gut feeling and a lack of counterexamples don’t constitute a proof. To prove that P is different from NP, you somehow have to rule out all potential algorithms for all of the hardest NP problems, a task that appears out of reach for current mathematical techniques. In fact, the field has coped by proving so-called barrier theorems, which say that entire categories of tempting proof strategies to resolve P versus NP cannot succeed. Not only have we failed to find a proof but we also have no clue what an eventual proof might look like.


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