First slide

Connected bodies

Question

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 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 A is sliding on the table, the tension in the string is

Moderate

A

$\frac{m_{1} m_{2} (1 + μ_{k}) g}{(m_{1} + m_{2})}$
B

$\frac{m_{1} m_{2} (1 - μ_{k}) g}{(m_{1} + m_{2})}$
C

$\frac{(m_{2} + μ_{k} m_{1}) g}{(m_{1} + m_{2})}$
D

$\frac{(m_{2} - μ_{k} m_{1}) g}{(m_{1} + m_{2})}$

Solution

Given situation is shown in the figure.
$\begin{array}{l} Here, N = m_{1} g \\ f = μ_{k} N = μ_{k} m_{1} g \end{array}$ . . . . .(i)
Let $α$ be the acceleration of blocks.
Equation of motion for A and B
$T - f = m_{1} a . . . . . (i) m_{2} g - T = m_{2} a . . . . . (ii)$
Adding equation
(ii) and (iii), we get.
$\begin{array}{l} m_{2} g - f = (m_{1} + m_{2}) a \\ a = \frac{m_{2} g - f}{m_{1} + m_{2}} \end{array}$
Put this value of $α$ in equation (iii)
$T = m_{2} g - m_{2} \frac{(m_{2} g - f)}{m_{1} + m_{2}}$
$= \frac{m_{1} m_{2} g + m_{1} m_{2} μ_{k} g}{m_{1} + m_{2}} = \frac{m_{1} m_{2} (1 + μ_{k}) g}{m_{1} + m_{2}}$ (using . (i))

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