Question

 A 50-gram block rests on an inclined plane 15^o with the horizontal. The coefficient of friction is 0.10. Find:  a) the acceleration of the block, b) the work done against friction 2 minutes after starting from rest, and c) its average power.

 

Answer 1

a) Let's apply the Newton's Second Law of Motion:

$$mgsin\theta-\mu mgcos\theta=ma,$$ $$a=g(sin\theta-\mu cos\theta),$$ $$a=9.8\ \dfrac{m}{s^2}\times(sin15^{\circ}-0.1\times cos15^{\circ})=1.59\ \dfrac{m}{s^2}.$$

b) Let's first find the final velocity of the block 2 minutes after starting from rest:

$$v=v_0+at=0+1.59\ \dfrac{m}{s^2}\times120\ s=190.8\ \dfrac{m}{s}.$$

Then, we can find the distance traveled by the block during this time:

$$d=\dfrac{1}{2}(v_0+v)t,$$ $$d=\dfrac{1}{2}\times(0+190.8\ \dfrac{m}{s})\times120\ s=11448\ m.$$

Finally, we can find the work done against friction 2 minutes after starting from rest:

$$W_{fr}=F_{fr}d=\mu mgdcos\theta,$$ $$W_{fr}=0.1\times0.05\ kg\times9.8\ \dfrac{m}{s^2}\times11448\ m\times cos15^{\circ}=542\ J.$$

c) By the definition of the power, we get:

$$P=\dfrac{W}{t}=\dfrac{542\ J}{120\ s}=4.52\ W.$$