Question

A block of mass $m$ sits on the top of a block of mass $2m$ which sits on a table. The coefficient of kinetic friction between all surface is $\mu =1$ . A massless string is connected to each mass and wraps halfway around a massless pulley, as shown. Assume that you pull on the pulley with a force of $6mg$ . The acceleration of your hand is -

• A. $\frac{g}{2}$
• B. $2g$
• C. $\frac{5g}{4}$
• D. $\frac{4g}{5}$

Answer 1

Answer: (C)

The free body diagrams are :

$F_1$ is force of friction between blocks
$F_2$ is force of friction between block and ground
Making FBD of pulley

Since pulley is light
$T=\frac{F}{2}=3mg$
Let blocks move together with common acceleration a
$a=\frac{6mg-F_2}{3m}=\frac{6mg-3mg}{3m}=g$
then for upper block
$T-F_1=ma\Rightarrow 3mg-F_1=mg\Rightarrow F_1=2mg$
but $F_1$ should be less than $f_{max}=\mu mg$
So, our assumption of no relative motion b/w blocks is incorrect that means there is relative motion
$\Rightarrow F_1$ is kinetic $=\mu mg$
$a_m=\frac{3mg-\mu mg}{m}2g\left(towardsright\right)$
$a_{2m}=\frac{3mg+\mu mg-3\mu mg}{2m}=\frac{g}{2}\left(towardsright\right)$
acceleration of pulley = accelerated of hand
$a_p=\frac{a_m+a_{2m}}{2}=\frac{2g+\frac{g}{2}}{2}=\frac{5g}{4}\left(towardsright\right)$