Question

A spring of spring constant $2N/m$ is attached to a mass of $1kg$ at one end and free at the other end. The system is allowed to free fall with the block below the spring. The tension force in the spring is
(Use $g = 10 m/{s^2}$)
(A) $0 N$
(B) $0.5 N$
(C) $1 N$
(D) $10 N$

Answer 1

Hint
Think that if you are pushing a block in a particular direction but you are unable to change its position, means displacement will be zero, then the force you exerted, has any magnitude or not. You can relate this case to the above question to find the solution.

Complete step by step answer
As per Hooke’s law, the force acting on an object by spring is always directly proportional to the spring's change in length away from its equilibrium length
The force exerted by a spring on an object attached to the one end of spring is directly proportional to the change in length of spring, as we know that when a spring expands its length is increased as compared to its normal length and when the spring is compressed its length decreases as compared to its normal length.
When we remove the sign of proportionality, we will get a spring constant.
$F = - kx$ (Here $k$ is a spring constant and $x$ is displacement)
In the above relation a negative sign shows that force and displacement always have opposite signs.
As per the question, one end of the spring is attached with a mass of $1kg$ and the other end is free fall, so there is no compression and expansion in the string's length, which means $x$ will be zero in this case and as per Hooke's law force is directly proportional to displacement. Then force exerted by spring on the block will also be zero.
$F = - k \times 0 = 0N$

Note
Point to be noted is, we should know that in the free fall case of spring the displacement is zero, according to Hooke’s law, force is directly proportional to displacement, so the force will also be zero in this case.