Question

A monkey pulls the midpoint of a $10cm$ long light inextensible string connecting two identical objects A and B of masses $0.3kg$ lying on smooth table continuously along the perpendicular bisector of line joining the masses. The masses are found to approach each other at a relative acceleration of $5m{s}^{-2}$ when they are $6cm$ apart. The constant force applied by monkey is:

• A. $4N$
• B. $2N$
• C. $3N$
• D. None of these

Answer 1

The correct option is B $2N$

With reference to the diagram
Note: The acceleration of both objects are symmetrical due to the symmetry of the system.
Relative acceleration $=5m/s^2$
$\Rightarrow 2a_1=5$
$\Rightarrow a_1=2.5m/s^2$
According to the free body diagram of left block,
$m{a}_1=Tcos\theta$
$\Rightarrow 0.3\times 2.5=\frac{3T}{5}$
$\Rightarrow T=1.25N$
According to free body diagram of the intersection point of force and string,
$F=2Tsin\theta$
$\Rightarrow F=2\times 1.25\times \left(\frac{4}{5}\right)$
$\Rightarrow F=2N$
Thus option (b) is the correct answer.