Energy & Work

AP Physics 1 review of work, kinetic and potential energy, the work-energy theorem, conservation of energy, and power.

# Energy & Work — AP Physics 1

Energy is one of the most powerful concepts in physics. The energy approach often lets you solve problems that would be extremely difficult with forces alone. AP Physics 1 expects you to understand work, kinetic energy, potential energy, conservation of energy, and power.

Key Concepts

Work

Work done by a constant force: W=FdcosθW = Fd\cos\theta where θ\theta is the angle between the force and displacement vectors.

  • Work is positive when force and displacement are in the same direction.
  • Work is negative when they are in opposite directions (e.g., friction).
  • Work is zero when force is perpendicular to displacement (e.g., normal force on a level surface).

Kinetic Energy

KE=12mv2KE = \frac{1}{2}mv^2

Work-Energy Theorem

Wnet=ΔKE=12mvf212mvi2W_{\text{net}} = \Delta KE = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Potential Energy

  • Gravitational PE: Ug=mghU_g = mgh (near Earth's surface)
  • Spring (Elastic) PE: Us=12kx2U_s = \frac{1}{2}kx^2

Conservation of Mechanical Energy

If only conservative forces do work: KEi+Ui=KEf+UfKE_i + U_i = KE_f + U_f

If non-conservative forces (like friction) act: KEi+Ui+Wnc=KEf+UfKE_i + U_i + W_{\text{nc}} = KE_f + U_f

Power

P=Wt=FvP = \frac{W}{t} = Fv Unit: watt (1 W=1 J/s1\ \text{W} = 1\ \text{J/s}).

Worked Example

Problem: A 2 kg2\ \text{kg} ball is released from rest at a height of 10 m10\ \text{m} on a frictionless ramp. What is its speed at the bottom?

Solution:

Using conservation of energy: mgh=12mv2mgh = \frac{1}{2}mv^2 v=2gh=2(9.8)(10)=196=14 m/sv = \sqrt{2gh} = \sqrt{2(9.8)(10)} = \sqrt{196} = 14\ \text{m/s}

Note: The mass cancels and the shape of the ramp doesn't matter — only the height.

Practice Questions

  1. 1. A 5 N5\ \text{N} force pushes a box 3 m3\ \text{m} along a surface at 60°60° to the horizontal. How much work is done?

    W=Fdcosθ=5(3)cos60°=7.5 JW = Fd\cos\theta = 5(3)\cos 60° = 7.5\ \text{J}.

    2. A spring with k=200 N/mk = 200\ \text{N/m} is compressed 0.1 m0.1\ \text{m}. How much elastic potential energy is stored?

    Us=12(200)(0.1)2=1 JU_s = \frac{1}{2}(200)(0.1)^2 = 1\ \text{J}.

    3. A 1500 kg1500\ \text{kg} car traveling at 20 m/s20\ \text{m/s} brakes to a stop. How much work does friction do?

    W=ΔKE=012(1500)(20)2=300,000 J=300 kJW = \Delta KE = 0 - \frac{1}{2}(1500)(20)^2 = -300{,}000\ \text{J} = -300\ \text{kJ}.

    4. An engine exerts 500 N500\ \text{N} on a car moving at 30 m/s30\ \text{m/s}. What power does it deliver?

    P=Fv=500×30=15,000 W=15 kWP = Fv = 500 \times 30 = 15{,}000\ \text{W} = 15\ \text{kW}.

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Summary

  • Work transfers energy to or from an object.
  • The work-energy theorem connects net work to the change in kinetic energy.
  • Conservation of energy is a powerful problem-solving tool — especially for systems with gravity and springs.
  • Power measures how quickly work is done.

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