Question

The system is released from rest and the block of mass $1\text{kg}$ is found to have a speed of $0.3\text{m/s}$ after it has descended through a distance of $1\text{m}$. The value of coefficient of kinetic friction is

• A. $0.24$
• B. $0.36$
• C. $0.12$
• D. $0.48$

Answer 1

The correct option is C $0.12$


As $1\text{kg}$ block is lowered by $1\text{m}$, pulley attached to it also gets lowered by $1\text{m}$. This releases $2\text{m}$ string length which is provided by displacement of $4\text{kg}$ block by $2\text{m}$ rightwards.
$(x{)}_{4\text{kg}}=2(x{)}_{1\text{kg}}$
$\Rightarrow (v{)}_{4\text{kg}}=2(v{)}_{1\text{kg}}$
$\therefore (v{)}_{4\text{kg}}=0.6\text{m/s}$
Speed of $4\text{kg}$ block will be $0.6\text{m/s}$ after descending through $h=1\text{m}$.

Applying Work Energy Theorem on system of blocks,
$W_{\text{ext}}=\Delta KE...\left(i\right)$
where $W_{\text{ext}}=W_f+W_{\text{gravity}}$
$W_f=-\mu Nd=-\mu (4\times 10\times 2)=-80\mu \text{J}$
$W_{gravity}=+mgh=+1\times 10\times 1=10\text{J}$
Substituting in Eq.$\left(i\right)$,
$W_{\text{friction}}+W_{\text{gravity}}={KE}_f-{KE}_i$
$\Rightarrow -80\mu +10=[\frac{1}{2}\times 4\times (0.6{)}^2+\frac{1}{2}\times 1\times \left(0.3{)}^2\right]-(0+0)$
$\Rightarrow -80\mu +10=\frac{18}{25}+\frac{9}{200}$
$\Rightarrow -80\mu =\frac{153}{200}-10$
$\Rightarrow -80\mu =\frac{-1847}{200}$
$\therefore \mu =0.115\simeq 0.12$
Coefficient of kinetic friction between the block and table is approximately $0.12$.