Notes
Work
- $W = Fd\cos\theta$, measured in joules. A force perpendicular to the motion does no work; friction does negative work.
Energy
- Kinetic energy $\displaystyle KE = \frac{1}{2}mv^2$. Work-energy theorem: $W_{\text{net}} = \Delta KE$.
- Gravitational potential energy $PE_g = mgh$. Elastic potential energy $\displaystyle PE_s = \frac{1}{2}kx^2$.
Conservation of energy
- Energy is never created or destroyed. Without friction, $KE_i + PE_i = KE_f + PE_f$.
- With friction, the "lost" mechanical energy becomes thermal energy.
- Falling from rest through height $h$: $\displaystyle v = \sqrt{2gh}$.
Power and efficiency
- Power $\displaystyle P = \frac{W}{t} = Fv$, in watts.
- Efficiency $\displaystyle = \frac{\text{useful output}}{\text{input}} \times 100\%$, always below 100% in real machines.
Topics in this unit
Each topic has its own short notes and practice.
Practice
$W = mgh = 500 \times 9.8 \times 12 = 58800 \text{ J}$.
$\displaystyle KE = \frac{1}{2}(0.15)(40^2) = 0.075 \times 1600 = 120 \text{ J}$.
$\displaystyle v = \sqrt{2(9.8)(10)} = \sqrt{196} = 14 \text{ m/s}$.
Energy is conserved; friction changes it to heat.
$W = Pt = 1500 \times 120 = 180000 \text{ J}$; remember to convert minutes to seconds.
$\displaystyle \frac{600}{800} = 0.75 = 75\%$.
Free response
Write your answer first, then open the scoring guide.
Free-response question
A 50 kg skier starts from rest at the top of a snowy hill that is 30 m high and skis to the bottom.
(a) What is the skier's gravitational potential energy at the top, using the bottom of the hill as zero height?
Show the answer and scoring guide
$PE = mgh = 50 \times 9.8 \times 30 = 14700 \text{ J}$.
Check off the points you earned:
(b) Ignoring friction and air resistance, how fast is the skier moving at the bottom?
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All the $PE$ becomes $KE$: $\displaystyle 14700 = \frac{1}{2}(50)v^2 \Rightarrow v^2 = 588 \Rightarrow v \approx 24.2 \text{ m/s}$.
Check off the points you earned:
(c) The skier's real speed at the bottom is $20 \text{ m/s}$. How much energy was changed by friction and air resistance, and what fraction of the starting energy remained as kinetic energy?
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$\displaystyle KE = \frac{1}{2}(50)(20^2) = 10000 \text{ J}$. Energy changed to heat $= 14700 - 10000 = 4700 \text{ J}$. Fraction remaining $\displaystyle = \frac{10000}{14700} \approx 68\%$.
Check off the points you earned: