Question

A block of mass $0.18kg$is attached to a spring of force constant $2N/{{m}^{-1}}$. The coefficient of friction between the block and the floor is $0.1$. Initially, the block is at rest and the spring is un-stretched. An impulse is given to the block as shown. The block slides for $0.6m$ and comes to rest. Initial velocity of block is $V=N/10$. Then value of N is

Answer 1

Hint: Given, the impulse is given to the block. If we conserve the energy of the block after impulse is given and it comes to rest, we will know the velocity as the kinetic energy is conserved. Also, include the energy produced by the frictional force and the spring force.
Formulas used:
$\begin{align}
  & K.E=\dfrac{1}{2}m{{v}^{2}} \\
 & {{E}_{f}}=\mu mgx \\
 & {{E}_{s}}= & \dfrac{1}{2}k{{x}^{2}}
\end{align}$

Complete answer:
Let the kinetic energy of the block initially be,
$K.E=\dfrac{1}{2}m{{v}_{1}}^{2}$
Then, the kinetic energy of the block finally be,
$K.E=\dfrac{1}{2}m{{v}_{2}}^{2}$
As it is given that the body comes to rest finally,
$\begin{align}
  & {{v}_{2}}=0 \\
 & K.E=0 \\
\end{align}$
Now, the energy produced due to friction is,
${{E}_{f}}=\mu mgx$
Energy produced due to the spring is,
$\begin{align}
  & {{E}_{s}}=\dfrac{1}{2}k{{x}^{2}} \\
 & \\
\end{align}$
Applying the conservation of energy,
$\begin{align}
  & \dfrac{1}{2}m{{v}^{2}}=\dfrac{1}{2}k{{x}^{2}}+\mu mgx \\
 & \dfrac{1}{2}(0.18){{(v)}^{2}}=\dfrac{1}{2}(2){{(0.6)}^{2}}+(0.1)(0.18)(10)\left( 0.6 \right) \\
 & v=0.4=\dfrac{4}{10} \\
\end{align}$
On comparing it with given question,
$N=4$

Additional information:
Elastic potential energy is the potential energy stored as a result of deformation of an elastic object, such as stretching of a spring. It is equal to work done to stretch the spring, which depends on the spring constant k as well as distance stretched. Taking Hooke’s law into consideration, force required is directly proportional to the amount of stretch. The calculation involving spring potential energy uses integral form.

Note:
The work done by the frictional force is negative in the above question. As the frictional force is actually in the L.H.S, we have written it in the R.H.S. if the spring is compressed, the distance travelled by the spring is taken as negative, whereas, if the spring is expanded, the distance travelled by the spring is taken as positive.