Question

A block of mass $m$ placed on a smooth floor is connected to a fixed support with the help of a spring of stiffness $k$. It is pulled by a rope as shown in the figure. Tension force $T$ of the rope is increased gradually without changing its direction until the block losses contact from the floor. The increase in rope tension $T$ is so gradual that acceleration in the block can be neglected. What is the extension in the spring, when the block losses contact from the floor?

• A. $\frac{mg\cos \theta }{k}$
• B. $\frac{mg\sin \theta }{k}$
• C. $\frac{mg\tan \theta }{k}$
• D. $\frac{mg\cot \theta }{k}$

Answer 1

The correct option is D $\frac{mg\cot \theta }{k}$

Free body diagram of the block, before it loses contact with the floor.

When the block is about to leave the floor, it is not pressing the floor.
Therefore $N=0$ and the block is in equilibrium.
$\sum F_x=0\Rightarrow T\cos \theta =kx\dots \dots \left(i\right)$
$\sum F_y=0\Rightarrow T\sin \theta =mg\text{(as N = 0)}\dots \dots \left(ii\right)$
From equation (i) and (ii), we have $x=\frac{mg\cot \theta }{k}$