Question

The spring is compressed by a distance ‘a’ and released. The block again comes to rest when the spring is elongated to a distance ‘b’. During this –
 

A) work done by spring on the block \[=\dfrac{1}{2}k{{(a+b)}^{2}}\]
B) work done by spring on the block \[=\dfrac{1}{2}k({{a}^{2}}-{{b}^{2}})\]
C) coefficient of friction \[=\dfrac{k(a-b)}{2mg}\]
D) coefficient of friction \[=\dfrac{k(a+b)}{2mg}\]

Answer 1

Hint: We can easily find the energy stored in each case of the spring while compressing and elongating. The algebraic sum of those quantities will give the sum total of the work done on the block. We can also find the coefficient of friction term.

Complete step-by-step solution:
We know that spring can store energy when it is either elongated or compressed. The block which is attached to the spring has this energy as work is done on it. The energy stored in the spring due to an external force will be the work done on the block.
 

 Let us find the energy store when the spring is compressed to a distance ‘a’ as –
\[\begin{align}
  & E=\dfrac{1}{2}k{{x}^{2}} \\
 & \text{here,} \\
 & x=-a \\
 & \Rightarrow E=\dfrac{1}{2}k{{a}^{2}}\text{ --(1)} \\
\end{align}\]
Now, the energy stored in the spring when the spring is elongated along the horizontal to a distance of ‘b’ units is given as –
\[\begin{align}
  & E=\dfrac{1}{2}k{{x}^{2}} \\
 & \text{here,} \\
 & x=+b \\
 & \Rightarrow E=\dfrac{1}{2}k{{b}^{2}}\text{ --(2)} \\
\end{align}\]
Now, we got the energy stored in the spring for both the cases. We already know that this energy stored is the work done on the block. The net work done on the block after compression and elongation is the difference between these two energies as they are in opposite directions. The total work done on the block is given as –
\[\begin{align}
  & W={{E}_{a}}-{{E}_{b}} \\
 & \Rightarrow W=\dfrac{1}{2}k{{a}^{2}}-\dfrac{1}{2}k{{b}^{2}} \\
 & \therefore W=\dfrac{1}{2}k({{a}^{2}}-{{b}^{2}}) \\
\end{align}\]
Thus, we get the total work done on the block.
Now, let us find the coefficient of friction involved in this situation for the block of mass ‘m’. We know that the total work done on the block is done against the frictional force acting opposite to its motion.
The frictional force is given as –
\[{{F}_{f}}=\mu mg\]
Where \[\mu \]is the coefficient of friction.
Now, the work done by the frictional force is given as –
\[\begin{align}
  & {{W}_{f}}=\mu mg(b)-\mu mg(-a) \\
 & \Rightarrow {{W}_{f}}=\mu mg(b+a) \\
\end{align}\]
Now, we can equate the work done on the block and the work done by the frictional force to get the coefficient of friction as –
\[\begin{align}
  & {{W}_{b}}={{W}_{f}} \\
 & \Rightarrow \dfrac{1}{2}k({{a}^{2}}-{{b}^{2}})=\mu mg(b+a) \\
 & \Rightarrow \mu =\dfrac{k(a+b)(a-b)}{2mg(a+b)} \\
 & \therefore \mu =\dfrac{k(a-b)}{2mg} \\
\end{align}\]
We get the coefficient of friction also.
These are the required answers.
The correct options are B and C.

Note: The coefficient of friction is a constant for a surface, it is a characteristic property of a material. In this case, we found the unknown through the known variables, where the unknown (\[\mu \]) is always a constant regardless of ‘a’, ‘b’, ‘k’ and ‘m’.