Question

In the figure given below, blocks $1$ and $3$ are connected by a belt and block $2$ rests on the horizontal portion of the belt. When the three blocks are released from rest, they accelerate with a magnitude of $0.5{\text{m/s}}^2$. Block 1 has mass $m$, block 2 has mass $2m$ and block 3 has the same mass $2m$. What is the coefficient of kinetic friction between block 2 and the belt?
[Take $g=9.81{\text{m/s}}^2$ and assume the mass of the belt to be negligible]

• A. 0.25
• B. 0.37
• C. 0.45
• D. 0.67

Answer 1

The correct option is B 0.37

Given that,
acceleration of the blocks $\left(a\right)=0.5{\text{m/s}}^2$
mass of block 1, $\left(m_1\right)=M$
mass of block 2, $\left(m_2\right)=2M$
mass of block 3, $\left(m_3\right)=2M$
FBD of the given system is


For block (1):
$T_{12}-mg=ma\dots \left(1\right)$
For block (2):
$T_{23}-2{\mu }_{k}mg-T_{12}=2ma\dots \left(2\right)$
For block (3)
$2mg-T_{23}=2ma\dots \left(3\right)$
Adding equation (1), (2) and (3),
$2mg-2{\mu }_{k}mg-mg=5ma$
$\Rightarrow g-2{\mu }_{k}g=5a$
$\Rightarrow {\mu }_k=\frac{g-5a}{2g}$
$\Rightarrow {\mu }_k=\frac{9.81-5\times 0.5}{2\times 9.81}$
$\therefore {\mu }_k=0.37$ is the coefficient of kinetic friction between block 2 and the table.