Answer to Question #126263 in Physics for Gia

Question

The graph shows power output as a function of time for the motor lifting the 18kg crate, initially at rest, up the 25* incline. The work done by the motor after 5.0s is 550J. If the final speed of the crate after 5.0 seconds is 2.0m/s, then the heat lost due to friction is_______J and the coefficient of friction is ____.

Answer 1

Assume that $a=const$.

a) $Q=W-KE-PE=550-\frac{mv^2}{2}-mg\Delta h$

$v=at\to a=v/t=2/5=0.4m/s^2$

$l=\frac{at^2}{2}=\frac{0.4\cdot5^2}{2}=5m \to$

$\Delta h=l\cdot \sin25°=5\cdot\sin25°=2.11m$

$Q=550-\frac{mv^2}{2}-mg\Delta h=550-\frac{18\cdot 2^2}{2}-18\cdot 9.81\cdot 2.11=141J$ Answer.

b) $W_{fr}=\mu mg\cos25°\cdot l \to \mu=\frac{W_{fr}}{mg\cos25°\cdot l}=$

$=\frac{Q}{mg\cos25°\cdot l}=\frac{141}{18\cdot 9.81 \cdot \cos25°\cdot 5}=0.177$ Answer.