Question

Consider the situation shown in figure. The system is released from rest and the block of mass 1.0 kg is found to have a speed 0.3 $\frac{m}{s}$ after it has descended through a distance of 1m. Find the coefficient of kinetic friction between the block and the table.

• A.
• B.
• C.
• D.

Answer 1

The correct option is A

Forces constraints we know that

$x_A=2x_B$

$\Rightarrow v_A=2v_B$

$\Rightarrow$ When speed of B is 0.3 m/s,speed of A will by 0.6 m/s

$\therefore \Delta KE=(\frac{1}{2}{m}_A{V}_A^2+\frac{1}{2}{M}_B{V}_B^2)-0$

$W_{net}$ = Work done gravity on block B

+ work done by friction on block A

+ work done by tension on block A

+ work done by tension on block B

= $M_{B}gx+(-\mu M_{A}g2x)+T_x+(-2T\frac{x}{2})$

= $M_{B}gx-2\mu M_{A}gx=\frac{1}{2}{M}_A{V}_A^2+\frac{1}{2}{M}_B{V}_B^2$

$\Rightarrow \frac{1}{2}\times 4\times (0.6{)}^2+\frac{1}{2}\times 1\times (0.3{)}^2=1\times 10\times 1-\mu \times 4\times 10\times 2$

$\Rightarrow \mu =0.12$