Power — Practice Problem Solutions

1. $W=mgh=(50)(10)(3.0)=1500\text{ J}$; $P=1500/5.0=300\text{ W}$. $P=300\text{ W}$.

2. $F=mg=(100)(10)=1000\text{ N}$ (constant speed); $P=Fv=(1000)(0.50)=500\text{ W}$. $P=500\text{ W}$.

3. $150{,}000\text{ W}\div746\text{ W/hp}\approx201\text{ hp}$. $\approx200\text{ hp}$.

4. $P=Fd/\Delta t$: A: $\dfrac{(180)(50)}{45}=200\text{ W}$. B: $\dfrac{(240)(30)}{30}=240\text{ W}$. C: $\dfrac{(210)(45)}{35}=270\text{ W}$. D: $\dfrac{(260)(35)}{40}=227.5\text{ W}$. Trial C has the greatest average power ($270\text{ W}$).

5. $P=mg\sin\theta,v$. $30°$: $(2.0)(10)(0.50)(2.0)=20\text{ W}$. $60°$: $(2.0)(10)(0.866)(2.0)\approx35\text{ W}$. $20\text{ W}$ and $\approx35\text{ W}$.

6. $a=2d/t^2=2(9.0)/9.0=2.0\text{ m/s}^2$; $F=ma=4.0\text{ N}$; $W=Fd=36\text{ J}$; $P=36/3.0=12\text{ W}$. Check: $\dfrac{2md^2}{t^3}=\dfrac{2(2.0)(81)}{27}=12\text{ W}$ ✓. $P_{avg}=12\text{ W}$.

7. $P_1=\dfrac{Fd}{t}$; $P_2=\dfrac{Fd}{t/2}=\dfrac{2Fd}{t}$. $P_2:P_1=2:1$. The work is identical; the second crate received it in half the time.

8. a) $\Delta K=\tfrac12(3.0)(4.0)^2=24\text{ J}$; $P_{avg}=24/2.0=12\text{ W}$. b) $a=4.0/2.0=2.0\text{ m/s}^2$; $F=ma=6.0\text{ N}$; $P=Fv=(6.0)(4.0)=24\text{ W}$. c) The force is constant but the speed grows from $0$ to $4.0\text{ m/s}$, so $P=Fv$ grows linearly from $0$ to $24\text{ W}$; the average of that linear rise is half of the final value, $12\text{ W}$.

9. $P_2=2P_1$. The same acceleration means the same net force ($F=ma$, same mass); $P=Fv$, and the speed doubled, so the power doubled.

10. (C) The area under a power-vs-time graph is the energy transferred: both trapezoids have area $\tfrac12(6+2)(4)=16\text{ J}$. The final power and the sign of the slope don't determine $\Delta E$.

11. (B) $d=\tfrac12at^2\Rightarrow a=\dfrac{2d}{t^2}$; $W=mad=\dfrac{2md^2}{t^2}$; $P_{avg}=\dfrac{W}{t}=\dfrac{2md^2}{t^3}$.