Work — Practice Problem Solutions

1. $W=Fd\cos0°=(15)(3.0)(1)=45\text{ J}$.

2. Zero work. Gravity points straight down while the displacement here is purely horizontal — the angle between them is $90°$, and $\cos90°=0$.

3. $\dfrac{W_1}{W_2}=\dfrac{\cos30°}{\cos60°}=\dfrac{\sqrt3/2}{1/2}=\sqrt3$. $W_1:W_2=\sqrt3:1$.

4. $W_{fan}=(0.5)(4.0)(-1)=-2\text{ J}$. $W_{gravity}=-mgh=-(0.5)(10)(1.0)=-5\text{ J}$. $W_{total}=-2+(-5)=-7\text{ J}$.

5. Triangle: $\tfrac12(3)(8)=12\text{ J}$. Rectangle: $(2)(8)=16\text{ J}$. Total: $28\text{ J}$.

6. a) Triangle: $\tfrac12(2)(1)=1.0\text{ J}$. b) Triangle + rectangle: $1.0+2.0=3.0\text{ J}$. c) $W_X:W_Y=1:3$.

7. (B) is correct. Net work equals the change in kinetic energy; the block's kinetic energy only changes while it's accelerating, so nonzero net work only happens then.