Question

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

• A. $\mu =0.115$
• B. $\mu =0.230$
• C. $\mu =0.145$
• D. $\mu =0.290$

Answer 1

Answer: (A)

$\mu =0.115$

From constraint relations, we can see that
$v_A=2v_B$
Therefore, $v_A=2\left(0.3\right)=0.6m/s$
as $v_B=0.3m/s$ (given)
Applying $W_{net}=\Delta U+\Delta K$

we get $-\mu m_{A}g{S}_A=-m_{B}g{S}_B+\frac{1}{2}{m}_A{v}_A^2+\frac{1}{2}{m}_B{v}_B^2$

Here, $S_A=2S_B=2mas{S}_B=1m$ (given)

$\therefore -\mu \left(4.0\right)\left(10\right)\left(2\right)=-\left(1\right)\left(10\right)\left(1\right)+\frac{1}{2}\left(4\right){\left(0.6\right)}^2+\frac{1}{2}\left(1\right){\left(0.3\right)}^2$

or $-80\mu =-10+0.72+0.045$
or $80\mu =9.235$
or $\mu =0.115$