Work, Energy, Power (Calculus-based) — AP Physics C: Mechanics Phys C Mech Study Guide

For: AP Physics C: Mechanics candidates sitting AP Physics C: Mechanics.

Covers: Line integral definition of work, work-energy theorem, conservative forces and potential energy, instantaneous power, and equilibrium of potential energy systems for AP Physics C Mechanics.

You should already know: Strong calculus (concurrent OK), AP Physics 1 helpful.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the AP Physics C: Mechanics style for educational use. They are not reproductions of past College Board papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official College Board mark schemes for grading conventions.

1. What Is Work, Energy, Power (Calculus-based)?

Work, energy, and power are interrelated scalar quantities that describe the transfer and storage of mechanical energy in systems, eliminating the need for complex vector force analysis for many motion problems. Unlike the algebra-based definitions taught in AP Physics 1, the AP Physics C framework uses integral calculus to handle variable forces, curved paths, and non-constant motion, aligning with CED Unit 3 requirements which make up 14-17% of your final exam score. This unit is one of the most frequently tested on both multiple choice and free response sections, as it connects to almost every other topic in the curriculum.

2. $W=\int \vec{F}\cdot d\vec{r}$

The line integral definition of work quantifies the energy transferred to or from a system by a force acting over a displacement, even when the force is not constant or the path of motion is curved.

Key Term Definitions:

• $\vec{F}$: Instantaneous force vector acting on the object at any point along the path
• $d\vec{r}$: Infinitesimal displacement vector tangent to the path of motion at that point
• Dot product: Ensures only the component of force parallel to displacement contributes to work: $\vec{F}\cdot d\vec{r}=Fdr\cos \theta$, where $\theta$ is the angle between $\vec{F}$ and $d\vec{r}$ at each point. For constant force and straight-line motion, this reduces to the familiar algebra formula $W=Fd\cos \theta$.

Worked Example

A variable force $\vec{F}(x)=(2x\hat{i}+3\hat{j})$ N acts on a particle moving along the x-axis from $x=0$ to $x=4$ m. Calculate the work done by the force:

1. The displacement vector is $d\vec{r}=dx\hat{i}$, so the dot product simplifies to $2xdx$ (the y-component of force is perpendicular to displacement, so it does zero work)
2. Integrate over the path: $$W={\int }_0^{4}2xdx={\left[x^2\right]}_0^4=16\text{ J}$$

Exam tip: Examiners frequently test this formula with springs, since spring force is variable: $F_{spring}=-kx$, so work done by a spring from displacement $x_1$ to $x_2$ is ${\int }_{x_1}^{x_2}-kxdx=\frac{1}{2}k{x}_1^2-\frac{1}{2}k{x}_2^2$.

3. Work-energy theorem

The work-energy theorem states that the net work done on an object by all external forces equals the change in its kinetic energy. This is valid for constant and variable forces, straight and curved paths, making it far more powerful than kinematic equations for most non-uniform motion problems.

Derivation (you may be asked to show this on FRQs):

1. Start with Newton's second law: $F_{net}=ma=m\frac{dv}{dt}$
2. Apply the chain rule to rewrite acceleration: $\frac{dv}{dt}=\frac{dv}{dr}\frac{dr}{dt}=v\frac{dv}{dr}$
3. Rearrange and integrate both sides: $${\int }_{r_i}^{r_f}{F}_{net}dr={\int }_{v_i}^{v_f}mvdv=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2=\Delta K$$

Worked Example

A 2 kg block is pushed by a variable force $F(x)=10-2x$ N along a frictionless horizontal surface from $x=0$ to $x=3$ m. If its initial velocity is 1 m/s, find its final velocity:

1. Calculate net work done: $$W_{net}={\int }_0^3(10-2x)dx={\left[10x-x^2\right]}_0^3=21\text{ J}$$
2. Set equal to change in kinetic energy: $21=\frac{1}{2}(2)(v_f^2-1^2)$
3. Solve for $v_f$: $v_f=\sqrt{22}\approx 4.69$ m/s

4. Conservative forces and potential energy

A conservative force is a force where the work done on an object moving between two points is independent of the path taken, and the total work done over a closed path (start and end at the same point) is zero. Examples include gravitational force, spring force, and electrostatic force. Non-conservative forces (friction, air resistance, applied motor force) have path-dependent work, so no potential energy function can be defined for them. For conservative forces, the work done equals the negative change in potential energy of the system: $$W_{cons}=-\Delta U=U_i-U_f$$ In integral form, potential energy is defined relative to a reference point where $U=0$ (e.g. ground for gravitational PE, spring equilibrium for elastic PE): $$U(r)=-{\int }_{r_{ref}}^r{\vec{F}}_{cons}\cdot d\vec{r}$$ You can also derive force from a potential energy function using the negative gradient: for 1D motion, $F_x=-\frac{dU}{dx}$.

Worked Example

The potential energy of a system is given by $U(x)=3x^2-2x^3$ J, where x is in meters. Find the conservative force acting on the system at $x=2$ m:

1. Take the first derivative of U: $\frac{dU}{dx}=6x-6x^2$
2. Apply the force-potential energy relation: $F_x=-\frac{dU}{dx}=-6x+6x^2$
3. Substitute $x=2$: $F=-12+24=12$ N

5. Power — instantaneous

Power is the rate at which work is done, or the rate of energy transfer between systems. Instantaneous power measures power at a specific moment in time, unlike average power which is total work divided by total time elapsed.

Derivation:

1. Start with the definition of power: $P=\frac{dW}{dt}$
2. Substitute $dW=\vec{F}\cdot d\vec{r}$: $$P=\vec{F}\cdot \frac{d\vec{r}}{dt}=\vec{F}\cdot \vec{v}$$ Instantaneous power is the dot product of the instantaneous force and instantaneous velocity vectors, with units of watts (1 W = 1 J/s). If force is at an angle $\theta$ to velocity, $P=Fv\cos \theta$: forces perpendicular to velocity deliver zero power (e.g. centripetal force does no work).

Worked Example

A 1000 kg car accelerates from rest at $2$ m/s² on a flat, frictionless road. What is the instantaneous power delivered by the car's engine at $t=5$ s?

1. Find velocity at $t=5$ s: $v=at=10$ m/s
2. Calculate engine force: $F=ma=2000$ N
3. Compute power: $P=Fv=2000\ast 10=20,000$ W = 20 kW

6. Equilibrium of potential energy

Equilibrium occurs where the net force on a system is zero. Since $F_x=-\frac{dU}{dx}$, equilibrium points are where the slope of the $U(x)$ graph is zero, i.e. $\frac{dU}{dx}=0$. There are three classifications of equilibrium:

1. Stable equilibrium: $U(x)$ is at a local minimum, $\frac{d^{2}U}{d{x}^2}>0$. A small displacement triggers a restoring force that pulls the system back to equilibrium.
2. Unstable equilibrium: $U(x)$ is at a local maximum, $\frac{d^{2}U}{d{x}^2}<0$. A small displacement triggers a force that pushes the system further from equilibrium.
3. Neutral equilibrium: $U(x)$ is constant over a region, $\frac{d^{2}U}{d{x}^2}=0$. A small displacement results in zero net force, so the system stays in its new position.

Worked Example

Given $U(x)=x^4-4x^2$ J, find all equilibrium points and classify them:

1. First derivative: $\frac{dU}{dx}=4x^3-8x$. Set to zero: $4x(x^2-2)=0$ → equilibrium at $x=0$, $x=\sqrt{2}$ m, $x=-\sqrt{2}$ m.
2. Second derivative: $\frac{d^{2}U}{d{x}^2}=12x^2-8$
3. Classify:

• $x=0$: $\frac{d^{2}U}{d{x}^2}=-8<0$ → unstable equilibrium
• $x=\sqrt{2}$: $\frac{d^{2}U}{d{x}^2}=16>0$ → stable equilibrium
• $x=-\sqrt{2}$: $\frac{d^{2}U}{d{x}^2}=16>0$ → stable equilibrium

7. Common Pitfalls (and how to avoid them)

• Wrong move: Using the algebra work formula $W=Fd$ for variable forces like springs. Why students do it: They carry over habits from AP Physics 1 and forget variable forces require integration. Correct move: Always use the line integral $\int \vec{F}\cdot d\vec{r}$ for any force that changes with position, time, or angle.
• Wrong move: Forgetting the negative sign in $W_{cons}=-\Delta U$. Why students do it: They mix up work done on the system vs work done by the system. Correct move: If a conservative force does positive work on an object, the object loses potential energy, hence the negative sign.
• Wrong move: Using average power instead of instantaneous power when asked for power at a specific time or position. Why students do it: They confuse $P_{avg}=W/\Delta t$ with $P_{inst}=\vec{F}\cdot \vec{v}$. Correct move: If the question uses phrases like "at t=3s" or "at x=5m", always use the dot product of instantaneous force and velocity.
• Wrong move: Counting potential energy as work done by non-conservative forces in the work-energy theorem. Why students do it: They don't separate conservative and non-conservative work. Correct move: Use the extended work-energy theorem: $W_{noncons}=\Delta K+\Delta U=\Delta E_{mech}$, so potential energy is separate from non-conservative work.
• Wrong move: Classifying equilibrium points using only the first derivative. Why students do it: They forget a zero first derivative only confirms an equilibrium point, not its type. Correct move: Always compute the second derivative or analyze the $U(x)$ curve shape to classify stable, unstable, or neutral equilibrium.

8. Practice Questions (AP Physics C: Mechanics Style)

Question 1 (10 points, FRQ style)

A force $\vec{F}(x)=(4x\hat{i}+5\hat{j})$ N acts on a 3 kg particle moving along the path defined by $y=2x$ from the origin (0,0) to the point (2, 4) m. (a) Calculate the total work done by the force on the particle. (b) If the particle starts from rest, what is its speed at (2,4) m?

Solution

(a) Express the displacement vector as $d\vec{r}=dx\hat{i}+dy\hat{j}$. Since $y=2x$, $dy=2dx$. Substitute into the work integral: $$W={\int }_0^2(4xdx+5\ast 2dx)={\int }_0^2(4x+10)dx={\left[2x^2+10x\right]}_0^2=28\text{ J}$$ (b) Apply the work-energy theorem: $W_{net}=\Delta K=\frac{1}{2}m{v}^2$. Substitute values: $$28=0.5\ast 3\ast v^2⟹v=\sqrt{\frac{56}{3}}\approx 4.32\text{ m/s}$$

Question 2 (5 points, MCQ style)

The potential energy function for a system is $U(x)=2x^3-9x^2+12x$ J, where x is in meters. Which of the following statements about equilibrium points is true? A) x=1 is unstable, x=2 is stable B) x=1 is stable, x=2 is unstable C) Both x=1 and x=2 are stable D) Both x=1 and x=2 are unstable

Solution

First find the first derivative: $\frac{dU}{dx}=6x^2-18x+12$. Set to zero: $6(x^2-3x+2)=0⟹x=1$ and $x=2$. Compute the second derivative: $\frac{d^{2}U}{d{x}^2}=12x-18$. At x=1: $\frac{d^{2}U}{d{x}^2}=-6<0$ (unstable). At x=2: $\frac{d^{2}U}{d{x}^2}=6>0$ (stable). Correct answer: A

Question 3 (7 points, FRQ style)

A 1500 kg car travels up a 10° incline at a constant speed of 20 m/s. The coefficient of kinetic friction between the tires and road is 0.2. Calculate the instantaneous power delivered by the car's engine.

Solution

Constant speed means net force is zero, so engine force equals the sum of gravitational force parallel to the incline and friction force:

1. Gravitational component: $mg\sin \theta =1500\ast 9.8\ast \sin (10°)\approx 2558$ N
2. Friction force: $f=\mu mg\cos \theta =0.2\ast 1500\ast 9.8\ast \cos (10°)\approx 2895$ N
3. Total engine force: $F=2558+2895=5453$ N
4. Instantaneous power: $P=Fv=5453\ast 20\approx 109,000$ W = 109 kW

9. Quick Reference Cheatsheet

10. What's Next

This unit is the foundation for all subsequent energy-based analysis in AP Physics C: Mechanics. Next, you will apply these concepts to systems of particles, collisions (where conservation of mechanical energy applies to elastic collisions), rotational motion (rotational kinetic energy, work done by torque), and gravitation (gravitational potential energy for orbital systems). Mastering the integral definitions and potential energy relations here will cut your problem-solving time in half for these later units, as energy analysis avoids complex vector calculations for force and acceleration.

If you struggle with any of the line integral calculations, equilibrium classification, or work-energy theorem applications, don't hesitate to ask Ollie, our AI tutor, for step-by-step help or additional practice problems tailored to your weak spots. You can also find more official College Board practice problems and full past exams on the homepage to test your readiness for the AP Physics C: Mechanics exam.