Question

In the arrangement shown in figure $m_A=4.0\text{kg}$ and $m_B=1.0\text{kg}.$ The system is released from rest and block $B$ is found to have a speed $0.3\text{m/s}$ after it has descended through a distance of $1\text{m}$. Find the coefficient of friction between the block $A$ and the table. Neglect friction elsewhere. (Take $g=10m/s^2$)

• A. $0.231$
• B. $0.501$
• C. $0.110$
• D. $0.23$

Answer 1

The correct option is C $0.110$


Let $x$ be displacement of A and $y$ be displacement of B
To find constraint relation:
$W_{net}^{Tension}=0$ (Since tension is an internal force)
$Tx-2Ty=0$
$x=2y$

Differentiating above equation
$\Rightarrow V_A=2V_B=2\left(0.3\right)=0.6{\text{ms}}^{-1}$
Given $y=1\text{m}\Rightarrow x=2\text{m}$
Applying work - Energy theorem,
$K_i+U_i+W_{f_k}=K_f+U_f$
$0+0+f_{k}xcos{180}^o=\frac{1}{2}{m}_A{V}_A^2+\frac{1}{2}{m}_B{V}_B^2-m_{B}gy$
$-\mu \left(40\right)\left(2\right)=\frac{1}{2}\left(4\right)(0.6{)}^2+\frac{1}{2}(1\left)\right(0.3{)}^2-\left(1\right)\left(10\right)\left(1\right)$
$-80\mu =0.72+0.45-10$
$-80\mu =-8.83$
$\mu =0.110$