AP Physics 1 Unit 3: Work, Energy, and Power | Notes and practice

Notes

Kinetic energy: $\displaystyle K = \frac{1}{2}mv^2$. Doubling speed makes $K$ four times bigger.
Work by a constant force: $W = Fd\cos\theta$, where $\theta$ is the angle between force and displacement. A force perpendicular to the motion does no work (like the normal force on a level floor).
Work-energy theorem: $W_{net} = \Delta K$.
On a force vs position graph, work is the area under the graph.

Gravitational: $\Delta U_g = mg\Delta h$ (near Earth's surface).
Spring: $\displaystyle U_s = \frac{1}{2}kx^2$.
Potential energy belongs to a system of interacting objects, like the Earth-ball system.

If only conservative forces (gravity, springs) do work inside the system, mechanical energy is conserved:

$$K_i + U_i = K_f + U_f$$

Friction converts mechanical energy into internal (thermal) energy: $\Delta E_{mech} = -f_k d$.
Energy bar charts help you track where energy goes.

$\displaystyle P = \frac{W}{\Delta t} = \frac{\Delta E}{\Delta t}$. Also $P = Fv$ for a constant force along the motion. Unit: watt (W = J/s).

How much work is done lifting a 2.0 kg box 3.0 m straight up at constant speed? ($g = 10$ m/s²)

A car's speed doubles. Its kinetic energy:

A waiter carries a tray horizontally at constant speed. How much work does the upward force on the tray do?

A 0.50 kg ball is dropped from 5.0 m. What is its speed just before hitting the ground (no air resistance, $g = 10$ m/s²)?

A spring with $k = 400$ N/m is compressed 0.10 m. How much energy does it store?

A motor does 1,200 J of work in 4.0 s. What is its power?

A box slides to a stop on a rough floor. Where did its kinetic energy go?

A net work of 50 J is done on a 4.0 kg object initially at rest. What is its final speed?

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Free-response question

A 60 kg skier starts from rest at the top of a 20 m high hill and slides down. Use $g = 10$ m/s².

(a) Ignoring friction, find the skier's speed at the bottom.

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$\displaystyle mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{2(10)(20)} = 20$ m/s.

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(b) The skier actually reaches the bottom at 16 m/s. How much mechanical energy was converted to thermal energy?

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Initial $U = 60(10)(20) = 12{,}000$ J. Final $\displaystyle K = \frac{1}{2}(60)(16)^2 = 7{,}680$ J. Thermal energy $= 4{,}320$ J.

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(c) If the hill were twice as long but the same height, would the final speed (with friction) be greater, less, or the same? Explain.

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Less. Friction acts over a longer distance, so it does more negative work and removes more mechanical energy.

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