Question

In the figure shown co-efficient of friction between the block B and the block C is 0.4. there is no friction between the block C and the surface on which it is placed. The system of blocks is released from rest in the situation shown in the diagram. Find the distance moved by the block C when block A descends through a distance 2 m. Given masses of the blocks are $m_A$ = 3 kg, $m_B$ = 5 kg and $m_C$ = 10 kg.

• A. 2 m
• B. 4 m
• C. 0 m
• D. 5 m

Answer 1

The correct option is A

2 m

Let there is no relative motion between the blocks B and C

Hence

T = $(m_B+m_C)a$ .........(1)

And

$m_{A}g-T=m_{A}a$ ..........(2)

from (1) and (2), we get

$a=\frac{m_{A}g}{m_A+m_a+m_c}=\frac{30}{18}=\frac{5}{3}m/s^2$

$\Rightarrow$ Net force on the block e C is,$F=m_{c}a=10\times \left(\frac{5}{3}\right)N=16.6N$

If maximum value of frictional force acting on block C is $f_{max}$ then

$F_{\left(max\right)}=\mu m_{B}g=0.4\times 10=20N\because F\le f_{max}$

Hence there is no relative motion between the block B and C. Therefore, distance moved by C is 2 m only.