Question

A disc of mass $50g$ slides with zero initial velocity down an inclined plane set at an angle ${30}^{\circ }$ to the horizontal. Having traversed a distance of $50cm$ along the horizontal plane, the disc stops. The work performed by the friction forces over the whole distance, assuming the friction coefficient $0.15$ for both inclined and horizontal planes is $(g=10m/s^2)$:

• A. $-0.05J$
• B. $0.5J$
• C. $-5J$
• D. $5J$

Answer 1

The correct option is A $-0.05J$

Retardation on horizontal surface,
$a_1=\mu g=0.15\times 10=1.5{m/s}^2$
Velocity just entering before horizontal surface,
$v=\sqrt{2a_1{S}_1}$
$=\sqrt{2\times 1.5\times 0.5}$
$=\sqrt{1.5}m/s$
Acceleration on inclined plane,
$a_2=g\sin \theta -\mu g\cos \theta$
$=10\times \frac{1}{2}-0.15\times 10\times \frac{\sqrt{3}}{2}$
$=3.7m/s^2$
$v=\sqrt{2a_2{S}_2}=\sqrt{1.5}$
$\therefore S_2=\frac{1.5}{2a_2}=\frac{1.5}{2\times 3.7}=0.2m$
$h=S_2\sin {30}^{\circ }=0.1m$
Now work done by friction is negative of initial mechanical energy.
$W=-mgh=-\left(0.05\right)\left(10\right)\left(0.1\right)$
$=-0.05J$