Question

A disc of mass $m=50g$ slides with the zero initial velocity down an inclined plane set at an angle $\alpha ={30}^{\circ }$ to the horizontal; having traversed the distance $l=50cm$ along the horizontal plane,the disc stops. If the work performed by the friction forces over the whole distance, assuming the friction coefficient $k=0.15$ for both inclined and horizontal planes is $\frac{x}{100}J$. Find $x$. (Write the answer to nearest integer)

Answer 1

The gain in kinetic energy of the disc when in reaches the bottom of the inclined plane

$=$ Loss in gravitational energy - Work done against friction

$=mgh-\mu mg\left(cos\alpha \right)\left(2h\right)=mgh[1-\sqrt{3}\mu ]$

This kinetic energy is lost by the frictional force when transversed horizontally

Loss in this kinetic energy $=\mu mgl$

Therefore, $\mu mgl=mgh[1-\sqrt{3}\mu ]$

$⟹h=\frac{\mu }{1-\sqrt{3}\mu }l$

Hence the total work done against friction $=\mu mgl+\sqrt{3}\mu mgh$

$=\mu mgl[1+\frac{\sqrt{3}\mu }{1-\sqrt{3}\mu }]=0.05J$