A block A of mass m 1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m 2 is suspended. The coefficient of kinetic friction between the block and the table is μ k . When the block B is sliding on the table, the tension in string is:

Question

A block A of mass m 1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m 2 is suspended. The coefficient of kinetic friction between the block and the table is μ k . When the block B is sliding on the table, the tension in string is:, m 2 + μ k m 1 g m 1 + m 2, m 2 - μ k m 1 g m 1 + m 2, m 1 m 2 1 + μ k g m 1 + m 2, m 1 m 2 1 - μ k g m 1 + m 2

MARKS App · Accepted Answer

From the figure, $\begin{aligned} & m_2 g-T=m_2 a \ldots \text { (i) } \ & T-\mu_k m_1 g=m_1 a \ldots \text { (ii) } \ & \text {multiply eqn (i) with } \mathrm{m}_1 \text { and eqn (ii) with } \mathrm{m}_2 \ & m_1 m_2 g-T m_1=m_1 m_2 a \ldots \text { (iii) } \ & T m_2-\mu_k m_1 m_2 g=m_1 m_2 a \ldots \text { (iv) } \ & \text{From eqn (iii) and (iv), we get,} \ & m_1 m_2 g-T m_1=T m_2-\mu_k m_1 m_2 g \ & m_1 m_2 g\left(1+\mu_k\right)=T\left(m_1+m_2\right) \ & \therefore T=\frac{m_1 m_2 g\left(1+\mu_k\right)}{m_1+m_2} \end{aligned}$

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