Question

A boy of mass $m$ climbs up a conveyor belt with a constant acceleration. The speed of the belt is $v=\sqrt{g h / 6}$ and the coefficient of friction between the boy and conveyor belt is $\mu=\frac{5}{3 \sqrt{3}}$. The boy starts from $A$ and moves with the maximum possible acceleration till he reaches the highest point $B$

Work done by friction on the boy is

• A. Equal to work done by boy
• B. Equal to work done by the motor in running the conveyor belt
• C. Zero
• D. None of above

Answer 1

Answer: (B)

$a =\frac{F_{\max }-m g \sin \theta}{m}$
$=\frac{\mu m g \cos \theta-m g \sin \theta}{m}=\frac{g}{3}$
$A B =v t+\frac{1}{2} a t^{2}$
$\frac{h}{\sin 30^{\circ}}=\sqrt{\frac{g h}{6}} t+\frac{1}{2} \frac{g}{3} t^{2}$
$g t^{2}+\sqrt{6 g h} t-12 h=0$
Solving, we get $t=\sqrt{6 \frac{h}{g}}$
$v_{f}=v_{i}+a t=\sqrt{\frac{g h}{6}}+\frac{g}{3} \sqrt{6 \frac{h}{g}}=3 \sqrt{\frac{g h}{6}}$
$\Delta KE =\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)=\frac{2}{3} m g h$
Distance moved by conveyor belt till the boy reaches $B$ :
$s=v_{i} t=\sqrt{\frac{g h}{6}} \sqrt{\frac{6 h}{g}}=h$
w.r.t. conveyor belt, distance moved by the boy,
$2 h-h=h$
Work done by gravity w.r.t. conveyor,
$W_{ mg }=-m g h \sin 30^{\circ}=-\frac{m g h}{2}$
From the frame of belt: $v_{i}=0, v_{f}=a t=2 \sqrt{\frac{g h}{6}}$
$W_{\text {boy }}=\Delta KE +\Delta PE$
$=\frac{1}{2} m\left(v_{f}^{2}-v_{i}^{2}\right)+m g \frac{h}{2}$
$=\frac{1}{2} m\left[4 \frac{g h}{6}\right]+\frac{m g h}{2}=\frac{5 m g h}{6}$
From the frame of ground:
$W_{\text {boy }}+W_{f}+W_{m g}=\Delta KE$
$W_{f}=\frac{2}{3} m g h+m g h-\frac{5}{6} m g h=\frac{5}{6} m g h$