The spring shown in figure (12-E5) is unstretched when a man starts pulling on the cord. The mass of the block is M. If the man exerts a constant force F. find (a) the amplitude and the time period of the motion of the block. (b) the energy stored in the spring when the block passes through the equilibrium position and (c) the kinetic energy of the block at this position

Hint We should know in Physics; energy is defined as the capacity of doing work. The existence of energy can be in the form of kinetic, thermal, electrical, chemical or nuclear and also in various other forms. Based on this concept we can solve this question.

Complete step by step answer
We should know that amplitude will be equal to the position when the force of the spring and the external force will be equal.
Let us consider that the amplitude is A.
So we can write that:
kA = F
So now, A = F/k
Here A is the amplitude
The total energy at any instant is energy to $\dfrac{{(k{A^2})}}{2}$
The energy that is stored in the block is the potential energy and is equal to $\dfrac{{(k{x^2})}}{2}$
Here x is the displacement from the mean position.
At the equilibrium x = 0
So, the answer of (b) is 0.
The kinetic energy is defined as the total energy and is equal to the $\dfrac{{(k{A^2})}}{2}$where A is F/ k
Therefore the kinetic energy is defined as $\dfrac{{(k{F^2})}}{{({k^2} \times 2)}}$

Note To solve such a question we should always keep in mind that kinetic energy is defined as the energy that is possessed by a body when it is motion and potential energy is the energy which is possessed by a body when it is in a state of rest.