Question

In the arrangement shown in figure, $m_A=4\text{kg}$ and $m_B=1\text{kg}$. The system is released from rest and block $B$ is found to have a speed of $0.3\text{m/s}$ after it has descended through a distance of $1\text{m}$. Find the coefficient of friction between the block and the table. Neglect friction elsewhere. (Take $g=10{\text{m/s}}^2$).

• A. $\mu =0.187$
• B. $\mu =0.115$
• C. $\mu =0.272$
• D. $\mu =0.323$

Answer 1

The correct option is B $\mu =0.115$

From constraint relations, we can say that
$V_A=2V_B$
Therefore, $V_A=2\left(0.3\right)\text{m/s}=0.6\text{m/s}$
Also if distance moved by $\text{B}$, $S_B=1\text{m}$
$S_A=2S_B=2\text{m}$

On applying work-energy theorem
$W_{non-cons}=\Delta U+\Delta K.E$
where the only non conservative force acting on the system is
friction $f=\mu m_{A}g$
We get: $-\mu m_{A}g{S}_A=-m_{B}g{S}_B+\frac{1}{2}{m}_A{V}_A^2+\frac{1}{2}{m}_B{V}_B^2$
$⟹-\mu \left(4\right)\left(10\right)\left(2\right)=-\left(1\right)\left(10\right)\left(1\right)+\frac{1}{2}\left(4\right)(0.6{)}^2+\frac{1}{2}(1\left)\right(0.3{)}^2$
$⟹(-80\mu )=(-10)+\left(0.72\right)+\left(0.045\right)$
$⟹80\mu =9.235$
$⟹\mu =0.115$