Question

Two blocks are connected by a string, as shown in the figure. They are released from rest. Find the speed after they have moved a distance x. $\mu$ is the coefficient of kinetic friction between the upper block and the surface. Assume that the pulley is massless and frictionless.

• A. $v=\sqrt{\frac{2gx{m}_2\mu }{(m_1+m_2)}}$
• B. $v=\sqrt{\frac{2g(m_2-m_1)x}{m_2}}$
• C. $v=\sqrt{\frac{2g(m_2-\mu m_1)x}{(m_1+m_2)}}$
• D. None of these

Answer 1

The correct option is C

$v=\sqrt{\frac{2g(m_2-\mu m_1)x}{(m_1+m_2)}}$

From the previous chapter we know that blocks will accelerate. Acquire a certain velocity after moving a certain distance.

Let's draw free body drawing of each block and apply the work energy theorem.

Block $m_1$ moves x meters to its right and acquires a velocity v

$W_N+W_T+W_{fr}+W_{gr}=\Delta K.E$.

$N.xcos{90}^{\circ }+Txcos{0}^{\circ }+\mu m_{1}gxcos{108}^{\circ }+m_{1}gcos{90}^{\circ }=\frac{1}{2}{m}_1{v}^2-\frac{1}{2}{m}_1{o}^2$

$Tx-\mu m_{1}gx=\frac{1}{2}{m}_1{v}^2$ -------------------(1)

$m_{2}gx-Tx=\frac{1}{2}{m}_2{v}^2$ ----------------------------(2)
(1)+(2) $\Rightarrow m_{2}gx-\mu m_{1}gx=\frac{1}{2}{v}^2(m_1+m_2)$
$\Rightarrow v=\sqrt{\frac{2gx(m_2-\mu m_1)}{m_1+m_2}}$