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 block's speed as a function of position $d$ down the incline is

• A. $\sqrt{\frac{4Mgd(\sin \theta +\mu \cos \theta )}{m+2M}}$
• B. $\sqrt{\frac{2Mgd(\sin \theta -\mu \cos \theta )}{m+2M}}$
• C. $\sqrt{\frac{4Mgd(\sin \theta -\mu \cos \theta )}{m+2M}}$
• D. $\sqrt{\frac{2Mgd(\sin \theta +\mu \cos \theta )}{m+2M}}$

Answer 1

The correct option is C $\sqrt{\frac{4Mgd(\sin \theta -\mu \cos \theta )}{m+2M}}$

From the isolated system model for the block-pulley-Earth system, using conservation of energy,
$$\begin{array}{c}\Delta K+\Delta U+\Delta E_{\text{int}}=0\\ \Rightarrow (\frac{1}{2}M{v}^2-0)+(\frac{1}{2}I{\omega }^2-0)+(0-Mgd\sin \theta )+fd=0\end{array}$$$\Rightarrow \frac{1}{2}M{v}^2+\frac{1}{2}\left(\frac{1}{2}m{r}^2\right){\left(\frac{v}{r}\right)}^2-Mgd\sin \theta +\left(\mu Mg\cos \theta \right)d=0$
$\Rightarrow v=\sqrt{\frac{4Mgd(\sin \theta -\mu \cos \theta )}{2M+m}}$