Question

When a mass $M$ is attached to the spring of force constant $k$, the spring stretches by $l$. If the mass oscillates with amplitude $l$, what will be the maximum potential energy stored in the spring?
A) $\dfrac{{kl}}{2}$
B) $2kl$
C) $\dfrac{1}{2}Mgl$
D) $Mgl$

Answer 1

Hint:
When a spring is stretched beyond its original length, a force must be given to it; this force relies on the spring constant and the length to which it has been stretched. Due to the length's expansion the potential energy will get stored inside the spring due to work done by this force.

Formula used:
The potential energy stored in the spring will be:
$U = \dfrac{1}{2}k{l^2} \\$

Complete step by step solution:
In order to know that a spring is stretched and extended by a certain amount of length when force is applied to it. $F = kx$ represents the force acting on the spring where $x$ is the spring's expanded extra length after being stretched, and $k$ is a constant.
According to the question, a mass of the spring$M$ is attached to the force which is constant $k$and the spring stretches by $l$. That means, $x = l$
Now, substitute the value of $x = l$in the above formula, then we have:
$F = kl$
Potential energy is the power that an object can store due to its position in relation to other things, internal tensions, electric charge, or other circumstances.
$U = \dfrac{1}{2}m{\omega ^2}{x^2}$
As we know, $Mg$force will act downward, so: $Mg = kl$
Therefore, the potential energy stored in the spring will be:
$U = \dfrac{1}{2}k{l^2} \\$
$\Rightarrow U = \dfrac{1}{2}kl \times l \\$
 $\Rightarrow U = \dfrac{{Mgl}}{2} \\$
Thus, the correct option is:(C) $\dfrac{1}{2}Mgl$






Note:
 It should be noted that stretching a spring results in the storage of potential energy. The work required to lengthen the spring is equivalent to potential energy. The force needed to extend a spring varies with distance, hence an integral is used in the work calculation.