Question

A 4 kg block is tied at one end of a massless belt as shown. The belt can slide on table. A 6 kg block is kept on the belt. The coefficient of friction between belt and table (${\mu }_1$) and between 6 kg block and belt (${\mu }_2$) is same and equal to 0.2. The system is released from rest. Answer the following questions: (Take g = 10 $m/s^2$)

Find the acceleration of 4 kg block

• A. zero
• B. 2
• C. 4
• D. 2.8

Answer 1

The correct option is C

4

To figure out if the 6 kg block slips on the belt. Let us assume, it does not, and the bodies move with a common acceleration a,

$\therefore$mg-T=ma (m=4kg) ...........(1)

T - $f_1-f_2=0$ (mass of belt is zero) ...........(2)

$f_2=ma$ (m=6kg) ...........(3)

As the belt will surely slip over the table

$f_1={\mu }_1\times N=0.2\times 6\times 10=12N$

$\therefore 40-T=4a$ ...........(1)

T-12-$f_2$=0 $\Rightarrow t-f_1=12$ ...........(2)

$f_2=6a$ ...........(3)

Adding (1) and (2)

40 - $f_2$=12+4a

Adding to (3)

40 = 12 + 10a

$\Rightarrow a=2.8m/s^2$

$\Rightarrow f_2=16.8N$

But $f_{2max}=\mu \times N$

$\therefore f_2$ cannot be 16.8 N, which means, the 6kg block slips over the belt

$\therefore f_2=f_{2max}=12N$

$\therefore$ equation become,

40 - T = 4a${}_1$ ...........(4)

T = $f_1+f_2$

T - $f_1-f_2$=0

$\Rightarrow T=f_1+f_2=12\times 1N=24N$

From (4)

40 - 24 = 4a

$\Rightarrow 16=4a_1$

$\Rightarrow a_1=4m/s^2$

Acceleration of 4 kg block is 4 $m/s^2$ downwards.