Question

Two 30 kg blocks rest on a massless belt which passes over a fixed pulley and is attached to a 40 kg block. If coefficient of friction between the belt and the table as well as between the belt and the blocks B & block C is $\mu$. The system is released from rest from the position shown, the speed with which the block B falls off the belt is

• A. $2\sqrt{2}m/s$ if $\mu =0.2$
• B. $2m/s$ if $\mu =0.2$
• C. $2\sqrt{5}m/s$ if $\mu =0.5$
• D. $2.5m/s$ if $\mu =0.5$

Answer 1

Answer: (A), (C)

The correct options are
A $2\sqrt{2}m/s$ if $\mu =0.2$
C $2\sqrt{5}m/s$ if $\mu =0.5$

Force acting on block B due after releasing the 40 kb block

Force on block B = friction friction force on block B

$F_B=\mu mg$

$=30\mu g$. . . . . (1)

Lets assume that the acceleration of block B is 'a' after releasing the 40 kg block.

We know that

Force = mass x acceleration

$F_B=ma$. . . . .(2)

Put, equation 1 = equation 2

$⟹ma=\mu mg$

$⟹a=\mu g$. . . . . (3)

We also know that

$V^2=2as$; where 's's is displacement

From the figure shown

Displacement of B will be 2 m when it is just about to fall.

$s=2$ m

Substitute 'a' from equation 3 and 's' from the above equation

$V=\sqrt{(}4\mu g)$

$⟹V=2\sqrt{(}\mu g)$

If $\mu =0.2$, then

$V=2\sqrt{(}2)\frac{m}{s^2}$

If $\mu =0.5$, then

$V=2\sqrt{(}5)\frac{m}{s^2}$

Therefore, Option A and Option C are correct.