Work and Kinetic Energy — AP Physics 1 Study Guide

For: AP Physics 1 candidates sitting AP Physics 1.

Covers: Definition of mechanical work, work done by constant and variable forces, kinetic energy calculation, the work-kinetic energy theorem, average and instantaneous power, and graphical work calculation for force vs. position graphs. Aligns with AP Physics 1 CED Unit 4 learning objectives.

You should already know: 1. How to resolve vectors into parallel and perpendicular components. 2. Newton's second law and basic kinematics for constant acceleration. 3. How to calculate area of simple geometric shapes (triangles, rectangles).

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 1 style for educational use. They are not reproductions of past College Board / Cambridge / IB papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official mark schemes for grading conventions.

1. What Is Work and Kinetic Energy?

Work and kinetic energy is the foundational topic of Unit 4 (Energy) in AP Physics 1, contributing approximately 4-6% of total exam weight per the official CED, and appears in both multiple-choice (MCQ) and free-response (FRQ) sections, often as a core component of larger energy-based problems. Work is defined as the transfer of mechanical energy between systems caused by a force acting over a displacement, while kinetic energy is the energy an object possesses due to its motion. This framework provides a much faster alternative to Newtonian kinematics for solving problems that relate speed and displacement, rather than speed and time. The core relationship between these two quantities, the work-energy theorem, is the starting point for all advanced energy concepts in the course, including potential energy and conservation of energy, which make up a much larger share of the exam.

2. Mechanical Work by a Constant Force

For a constant force $\vec{F}$ acting on an object that undergoes a displacement $\vec{d}$, work $W$ (a scalar quantity) is the dot product of force and displacement: $$W=\vec{F}\cdot \vec{d}=Fd\cos \theta$$ where $\theta$ is the angle between the force vector and the displacement vector, and the SI unit of work is the joule ($1\text{J}=1\text{N}\cdot \text{m}$). The key insight here is that only the component of force parallel to displacement does work: the perpendicular component contributes zero work because $\cos 90^{\circ }=0$. Work can be positive or negative: if $\theta <90^{\circ }$, work adds energy to the object; if $\theta >90^{\circ }$, work removes energy from the object. Friction and drag almost always do negative work, since they act opposite the direction of motion. For multiple forces, net work is the algebraic sum of work done by each individual force.

Worked Example

A student pulls a 12 kg sled across frictionless horizontal ice with a rope that makes a 30° angle with the horizontal. Tension in the rope is 40 N, and the sled moves 5.0 m horizontally. What is the work done on the sled by tension?

1. Identify given values: $F_T=40\text{N}$, $\theta =30^{\circ }$, $d=5.0\text{m}$. Gravity and normal force are perpendicular to displacement, so they do zero work, so we only calculate work from tension.
2. Apply the constant-force work formula: $W=Fd\cos \theta$.
3. Substitute values: $W=(40)(5.0)\cos (30^{\circ })=200(0.866)=173.2\text{J}$.
4. Round to two significant figures (consistent with given values): $W=170\text{J}$.

Exam tip: Always confirm which axis the problem's given angle is measured from. If the angle is given from the vertical instead of the horizontal, use $\sin \theta$ instead of $\cos \theta$ to get the parallel component of force.

3. Kinetic Energy and the Work-Energy Theorem

Kinetic energy ($K$) is the energy of an object's motion, defined for an object of mass $m$ moving at speed $v$ as: $$K=\frac{1}{2}m{v}^2$$ Kinetic energy is always non-negative (since mass is positive and speed squared is positive) and is a scalar, so it has no direction, only magnitude. The work-energy theorem, derived directly from Newton's second law, states that the net work done on an object by all forces acting on it equals the change in the object's kinetic energy: $$W_{\text{net}}=\Delta K=K_f-K_i=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$$ This theorem works for both constant and variable forces, making it the go-to tool for solving problems that ask for final speed after a known displacement, without needing to calculate acceleration first.

Worked Example

Using the sled from the previous example, if the sled starts from rest, what is its speed after moving 5.0 m?

1. Net work equals work done by tension, since friction is zero and all other forces do zero work: $W_{\text{net}}=173\text{J}$.
2. Initial kinetic energy $K_i=0$, since the sled starts from rest.
3. Apply the work-energy theorem: $W_{\text{net}}=\frac{1}{2}m{v}_f^2-0$.
4. Rearrange and solve for $v_f$: $v_f=\sqrt{\frac{2W_{\text{net}}}{m}}=\sqrt{\frac{2(173)}{12}}=\sqrt{28.8}\approx 5.4\text{m/s}$.

Exam tip: If a problem gives displacement and asks for speed, always check if the work-energy theorem is faster than kinematics. It will save you 1-2 minutes on most MCQ questions.

4. Work Done by a Variable Force

When force is not constant (e.g., the force from a stretched spring, or a changing applied force), the constant-force work formula does not apply. For AP Physics 1, you only need to remember that work done by a variable force is equal to the total area under a force vs. position ($F$ vs. $x$) graph between the starting and final position. Area above the x-axis counts as positive work, and area below the x-axis counts as negative work. You can calculate total work by splitting the graph into simple geometric shapes (triangles, rectangles, trapezoids), calculating the area of each, then adding them with the correct sign. This is one of the most common MCQ topics for work and energy on the AP exam.

Worked Example

A variable force acts on a 2 kg cart moving along the x-axis, with the force profile given as: $F$ increases linearly from 0 N at $x=0$ to 8 N at $x=2$ m, stays constant at 8 N from $x=2$ m to $x=5$ m, then decreases linearly back to 0 N at $x=7$ m. What is the total work done by the force between $x=0$ and $x=7$ m?

1. Split the graph into three regions to calculate area: 0-2 m, 2-5 m, 5-7 m.
2. Calculate area for each region: Region 1 (0-2 m) is a triangle: $W_1=\frac{1}{2}\times 2\text{m}\times 8\text{N}=8\text{J}$. Region 2 (2-5 m) is a rectangle: $W_2=(5-2)\text{m}\times 8\text{N}=24\text{J}$. Region 3 (5-7 m) is a triangle: $W_3=\frac{1}{2}\times 2\text{m}\times 8\text{N}=8\text{J}$.
3. Sum the areas: $W_{\text{total}}=8+24+8=40\text{J}$.

Exam tip: If the force crosses from positive to negative on the $F-x$ graph, don't forget to subtract the area of the negative region, don't just add all areas regardless of sign.

5. Power

Power is the rate at which work is done, or energy is transferred, between systems. Average power over a time interval $\Delta t$ is given by: $$P_{\text{avg}}=\frac{W}{\Delta t}$$ The SI unit of power is the watt ($1\text{W}=1\text{J/s}$). For instantaneous power (the power at a specific moment in time), if a force $F$ is parallel to the object's velocity $v$, power simplifies to $P=Fv$. This is useful for problems involving vehicles, engines, or human movement, where power output is often given, and you need to solve for time, force, or speed.

Worked Example

A 1500 kg car accelerates from rest to 20 m/s, with negligible friction. If the car's engine delivers an average power of 40 kW, how much time does the acceleration take?

1. The change in kinetic energy of the car equals the net work done by the engine: $W=\Delta K=\frac{1}{2}m{v}_f^2-0=0.5(1500)(20{)}^2=300,000\text{J}=300\text{kJ}$.
2. Convert average power to watts: $40\text{kW}=40,000\text{W}$.
3. Rearrange the power formula to solve for time: $\Delta t=\frac{W}{P_{\text{avg}}}=\frac{300000}{40000}=7.5\text{s}$.

Exam tip: Always convert kilowatts to watts (multiply by 1000) before calculating energy or time. A common mistake leaves power in kilowatts and gets a time 1000 times smaller than the correct value.

6. Common Pitfalls (and how to avoid them)

• Wrong move: Using the full magnitude of an angled force in the work formula, omitting the $\cos \theta$ term. Why: Students memorize the simplified $W=Fd$ for parallel forces and forget to adjust for angled forces. Correct move: Always resolve the force into parallel and perpendicular components, and only use the parallel component for work calculations.
• Wrong move: Plugging work done by a single force into the work-energy theorem instead of net work. Why: Students confuse "work done by the applied force" with "net work from all forces". Correct move: Always sum work done by every force (including friction, gravity, and normal force) to get $W_{\text{net}}$ before applying the theorem.
• Wrong move: Treating kinetic energy as a vector and adding it via vector components. Why: Students are used to working with velocity and force vectors, so they carry over vector addition to kinetic energy. Correct move: Remember kinetic energy is a scalar; add magnitudes directly, no components required.
• Wrong move: Counting area below the x-axis on a $F-x$ graph as positive work. Why: Students remember "area equals work" but forget force direction changes the sign. Correct move: Always assign a negative sign to area below the x-axis when calculating total work.
• Wrong move: Using $P=Fv$ to calculate average power for an accelerating object. Why: Students memorize the simplified power formula and use it for any problem. Correct move: Only use $P=Fv$ for instantaneous power; use $P=W/\Delta t$ for average power over a time interval.

7. Practice Questions (AP Physics 1 Style)

Question 1 (Multiple Choice)

A 0.5 kg ball is thrown straight upward with an initial speed of 10 m/s. Air resistance does -4 J of work on the ball as it rises to its maximum height. What is the maximum height of the ball relative to the release point? A) 4.1 m B) 5.1 m C) 6.2 m D) 9.8 m

Worked Solution: Apply the work-energy theorem. At maximum height, final speed is 0, so $\Delta K=0-\frac{1}{2}m{v}_i^2=-0.5(0.5)(10{)}^2=-25\text{J}$. Net work equals work done by gravity plus work done by air resistance: $W_{\text{net}}=-mgh-4\text{J}$. Set equal to $\Delta K$: $-mgh-4=-25$, so $mgh=21$. Solve for $h$: $h=21/(0.5\times 9.8)=21/4.9\approx 4.1\text{m}$. If you forgot to subtract work done by air resistance, you get 5.1 m, the most common distractor. The correct answer is A.

Question 2 (Free Response)

A 50 kg crate is pushed across a rough horizontal floor by a constant 200 N horizontal force. The coefficient of kinetic friction between the crate and floor is 0.3. The crate starts from rest and moves 8.0 m. (a) Calculate the work done by the pushing force, work done by friction, and net work on the crate. (b) Use the work-energy theorem to find the final speed of the crate. (c) If the same 200 N force is applied at a 20° angle below the horizontal instead of horizontally, will the final speed after 8.0 m be higher, lower, or the same than in part (b)? Justify your answer.

Worked Solution: (a) Work done by the pushing force: $W_F=Fd=(200\text{N})(8.0\text{m})=1600\text{J}$. Normal force equals weight: $N=mg=50(9.8)=490\text{N}$. Friction force: $f_k={\mu }_{k}N=0.3(490)=147\text{N}$. Work done by friction: $W_f=-f_{k}d=-147(8)=-1176\text{J}$. Gravity and normal force do zero work, so net work: $W_{\text{net}}=1600-1176=424\text{J}$. (b) Work-energy theorem: $W_{\text{net}}=\frac{1}{2}m{v}_f^2-0$, so $v_f=\sqrt{\frac{2W_{\text{net}}}{m}}=\sqrt{\frac{2(424)}{50}}\approx 4.1\text{m/s}$. (c) The final speed will be lower. Pushing at an angle below the horizontal adds a downward component to the applied force, which increases the normal force on the crate ($N=mg+F\sin \theta$ instead of $N=mg$). Higher normal force increases kinetic friction, which increases the magnitude of negative work done by friction. This reduces net work, which reduces the final kinetic energy and final speed.

Question 3 (Application / Real-World Style)

A cyclist accelerates from 5 m/s to 10 m/s on a flat road. The total mass of the cyclist and bicycle is 80 kg. The cyclist's legs deliver an average power of 420 W over the 10 second acceleration. Friction and air resistance do -1200 J of work during the acceleration. What is the efficiency of the cyclist (efficiency is the percentage of work done by the legs that goes into increasing kinetic energy)?

Worked Solution: First, calculate total work done by the cyclist's legs: $W=P_{\text{avg}}\Delta t=420\text{W}\times 10\text{s}=4200\text{J}$. Next, calculate the change in kinetic energy of the system: $\Delta K=\frac{1}{2}m(v_f^2-v_i^2)=0.5(80)(10^2-5^2)=3000\text{J}$. Efficiency is $\left(\frac{\text{useful energy}}{\text{total input energy}}\right)\times 100\%=\left(\frac{3000}{4200}\right)\times 100\%\approx 71\%$. In context, this means 71% of the work done by the cyclist's legs goes into increasing kinetic energy, with the remaining 29% lost to overcoming friction and air resistance.

8. Quick Reference Cheatsheet

9. What's Next

Work and kinetic energy is the fundamental prerequisite for all remaining energy topics in AP Physics 1 Unit 4. Next, you will extend the work-energy theorem to include potential energy, stored energy due to position, leading to the principle of conservation of energy for closed systems. Without mastering how to calculate work and apply the work-energy theorem, you will not be able to correctly solve problems involving gravitational or elastic potential energy, which make up the majority of energy-related FRQ questions on the AP exam. This topic also connects directly to power in electric circuits later in the course, where power is defined identically as the rate of energy transfer. Continue your study with these next topics:

Potential Energy Conservation of Energy Power for Energy Systems