Question

An object of mass M is attached to a vertical spring which slowly lowers to equilibrium position. This extends the spring by x. If the same object is attached to the same vertical spring but permitted to fall suddenly instead, then the extension of the spring is
A. $\dfrac{x}{2}$
B. $2x$
C. $3x$
D. $x$

Answer 1

Hint: First equate mg (gravity) and kx (spring force)and then calculate the value of x. In the second case, use work energy theorem i.e. W=$\Delta $K. Work done by gravity is ${{W}_{g}}=Mgx'$ and work done by spring force is ${{W}_{s}}=-\dfrac{1}{2}kx{{'}^{2}}$. And $\Delta $K=0.

Formula used:
${{F}_{g}}=Mg$
${{F}_{s}}=-kx$
${{W}_{s}}=\dfrac{1}{2}kx{{'}^{2}}$
${{W}_{g}}=Mgx'$
W=$\Delta $K

Complete step by step answer:
It is given that an object of mass M is attached to a spring. Let the spring constant of this spring be k. When there is change
There are two cases in the question. In the first case, the object is brought down slowly and in the second case, the object is left suddenly.
In the first case, the object is lowered by a distance of x and the object is at rest at this position. Therefore, the spring force balances the gravitational force. Hence, we get
$Mg=kx$
$\Rightarrow x=\dfrac{Mg}{k}$
In the second case, when the spring is left suddenly the spring stretches as the gravitational force will pull it downwards. Along with the gravitational, the spring will pull the object upwards. As the extension in the spring increases, the spring force will also increase because spring force is equal to kx.
There will come a time when the object will come to rest. Let the extension in spring at this time be x’. In this process, the work done by the gravitational force is ${{W}_{g}}=Mgx'$.
Work done on the object by the spring is given as ${{W}_{s}}=\dfrac{1}{2}k{{x}^{2}}$.
In this case, the direction of the spring force and displacement of the object are anti-parallel (opposite in direction). Thus, the spring does negative work. Therefore, the work done by spring in this case is
 ${{W}_{s}}=-\dfrac{1}{2}kx{{'}^{2}}$.
Since the object was at rest initially and is at rest now, the change in kinetic energy of the object is zero. According to the work-energy theorem, the total work done is equal to the change in kinetic energy of the body.
Therefore, ${{W}_{g}}+{{W}_{s}}=0$
$Mgx'-\dfrac{1}{2}kx{{'}^{2}}=0$
$\Rightarrow x'\left( Mg-\dfrac{1}{2}kx' \right)=0$
Therefore, x’=0 or $Mg-\dfrac{1}{2}kx'=0$.
We will discard x’=0 because we know that the extension in the spring is not zero.
Therefore, $Mg-\dfrac{1}{2}kx'=0$
$\Rightarrow Mg=\dfrac{1}{2}kx'$
$\Rightarrow x'=2\dfrac{Mg}{k}$
We found that $x=\dfrac{Mg}{k}$.
Therefore, $\Rightarrow x'=2x$
Therefore, the extension in the spring when it is left suddenly is equal to 2x.
Hence, the correct option is b.

Note: If one is thinking what is the difference between leaving the object slowly and leaving the object suddenly.
We already know what happens when the object is left suddenly.
When the object is lowered down slowly, maybe with the help of our hand, along with the force exerted by the spring and the gravitational force, our hand also exerts an equal force on the object in such a way that the acceleration of the object is almost zero. All the three forces are balanced and the net force on the object is zero.
Initially when the spring is in its original length, the gravitational force pulls it downwards and the spring force is less than the gravitational force. However the force exerted by our hand helps to balance the gravitational force. As the object comes down, the spring force increases and we slowly decrease the force exerted by our hand. When the object reaches the equilibrium, i.e. when the gravitational force is equal to the spring force, we remove our and the object stays at rest.