Question

A 4kg roller is attached to a massless spring of spring constant $k=100N{{m}^{-1}}$. It rolls without slipping along a frictionless horizontal road. The roller is displaced from its equilibrium position by 10cm and then released. Its maximum speed will be
A. $0.5m{{s}^{-1}}$
B. $0.6m{{s}^{-1}}$
C. $0.4m{{s}^{-1}}$
D. $0.8m{{s}^{-1}}$

Answer 1

Hint: Recall the expression for potential energy of the displaced spring. Also, find out the expression for kinetic energy gained when the spring is being released from the maximum displaced position. Apply law of conservation of energy to get the expression for maximum speed and substitute the given values to find the maximum speed.
Formula used:
Expression for potential energy of a spring,
$P.E=-\dfrac{k{{x}_{m}}^{2}}{2}$
Expression for kinetic energy,
$K.E=\dfrac{1}{2}m{{v}^{2}}$

Complete answer:
In the question, we are given a 4kg roller that is attached to a massless spring. This roller is rolling without slipping on a frictionless horizontal road. Now, if the roller is displaced from its equilibrium position by 10cm and released, we are asked to find its maximum speed. We are also given the value of spring constant as,
$k=100N{{m}^{-1}}$

From the force law for the spring which is also called Hooke’s law, we know that,
${{F}_{s}}=-kx$
Also the work done by the spring force can be given by,
${{W}_{s}}=\int\limits_{0}^{{{x}_{m}}}{{{F}_{s}}}dx=-\int\limits_{0}^{{{x}_{m}}}{kxdx}$
$\Rightarrow {{W}_{s}}=-\dfrac{k{{x}_{m}}^{2}}{2}$
This work done is being stored as potential energy in the roller when it is displaced from the equilibrium position. Therefore,
$P.E=-\dfrac{k{{x}_{m}}^{2}}{2}$ ……………………………………….. (1)
We have been told that the roller is being released from its displaced position. So we know that the potential energy stored while the roller is in displaced position is transferred to kinetic energy when released.
We know that, the expression for Kinetic energy is given by,
$K.E=\dfrac{1}{2}m{{v}^{2}}$ …………………………………. (2)
Also, from the law of conservation of energy, we know that the potential energy is being transferred as the Kinetic energy in the roller. So we could equate (1) and (2) to get,
$\Rightarrow \dfrac{1}{2}k{{x}_{m}}^{2}=\dfrac{1}{2}m{{v}^{2}}$
$\Rightarrow k{{x}_{m}}^{2}=m{{v}^{2}}$
$\Rightarrow v=\sqrt{\dfrac{k{{x}_{m}}^{2}}{m}}$
Now, we could directly substitute the given values to get,
$\Rightarrow v=\sqrt{\dfrac{100N{{m}^{-1}}\times {{\left( 0.1m \right)}^{2}}}{4}}$
$\Rightarrow v=\dfrac{10\times 0.1}{2}=0.5m{{s}^{-1}}$
Since we have considered the maximum displaced position for the roller ${{x}_{m}}$, the speed v that we got is the maximum speed.
Therefore, the maximum speed of the roller will be $0.5m{{s}^{-1}}$.

So, the correct answer is “Option A”.

Note:
While doing numerical problems, make sure that you convert all the given values in their SI units. For example here the displacement of the spring is given in cm, so, we have converted it into meters before substitution. All the other values are in their SI units and the options are also in the SI unit of speed.