A block A of mass `m_1` rests on a horizontal table. A light string connected to it passes over a frictionless pully at the edge of the table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is H. When the block A is sliding on the table, the tension in the string is

Options

A. `(m_1 m_2(1+mu _k ) g)/( (m_1 +m_2))`
B. `(m_1m_2 (1- mu_k)g)/((m_1) +m_2)`
C. `((m_2 +mu_k m_1)g)/((m_1 +m_2))`
D. `((m_2 -mu_k m_1 ) g)/((m_1 +m_2))`

Answer

A

Explanation

Given situation is shown in the figure
here `,N= m_1 g ,f = mu_k N = mu_k m_1 g `…(i)
Let a be the acceleation on blocks .
equation of motion for A and B

` T-f = m_1 a … (ii) `
`m_2 g - T =m_2 a …(iii)`
Adding equation (ii) and (iii)
we get
` m_2 g - f = (m_1 +m_2) a`
` a =(m_2 g-f)/( m_1 +m_2)`
substiting value of a in equation (iii)
` T = m_2 g -m_2 ((m_2 g -f))/(m_1 +m_2)`
`=(m_1 m_2 g + m_1 m_2 mu_s g)/(m_1 +m_2) = (m_1 m_2 (1 + mu _s ) g)/( m_1 +m_2)` ( using (i))