Question

A cord is wrapped around a pulley that is shaped like a disk of mass $m$ and radius $r$. The cord's free end is connected to a block of mass $M$. The block starts from rest and then slides down an incline that makes an angle $\theta$ with the horizontal as shown in figure. The coefficient of kinetic friction between block and incline is $\mu$ . Then the magnitude of the acceleration of the block is

• A. $\frac{2Mg(\sin \theta -\mu \cos \theta )}{2M+m}$
• B. $\frac{2Mg(\sin \theta +\mu \cos \theta )}{2M+m}$
• C. $\frac{2Mg(\sin \theta -\mu \cos \theta )}{2M-m}$
• D. $\frac{Mg(\sin \theta -\mu \cos \theta )}{2M+m}$

Answer 1

The correct option is A $\frac{2Mg(\sin \theta -\mu \cos \theta )}{2M+m}$


From the FBD of block of mass $M$,
$Mgsin\theta -T-\mu Mgcos\theta =Ma$ ......(1)
From the FBD of pulley,
Torque, $T^{\prime }=T\times r=I\times \alpha$
$T\times r=\frac{m{r}^2}{2}\alpha$
$\because a=\alpha \times r$
$\therefore T=\frac{ma}{2}$ ........(2)

Putting value of $T$ in equation (1),
we get, $Mgsin\theta -\frac{ma}{2}-\mu Mgcos\theta =Ma\Rightarrow a=\frac{2Mg(\sin \theta -\mu \cos \theta )}{2M+m}$