Question

If the force constant of a wire is K, the work done in increasing the length l of the wire by l is
A. $3Kl$
B. $2Kl$
C. $\dfrac{{K{l^2}}}{2}$
D. $\dfrac{{K{l^2}}}{3}$

Answer 1

Hint: When some amount of work is done against the conservative force then the work done will be stored in the system as the potential energy as energy can neither be created nor destroyed. By using this principle we will solve this question.

Formula used:
$\eqalign{
  & F = - Kx \cr
  & U = \dfrac{1}{2}K{x^2} \cr} $

Complete step-by-step answer:
We can compare this wire with the spring. In spring, as long as spring is in natural length it doesn’t have any energy. If we apply force to elongate it or compress it then the elongation or compression is produced in it and potential energy will be stored in the spring.
The spring force which opposes the applied force will be $F = - Kx$
Where ‘K’ is the force constant and ‘x’ is the elongation or compression produced and the negative sign indicates that spring force opposes the direction of change.
Initially for the wire there is no energy. The amount of work done to increase its length will be stored as internal potential energy.
Initial energy = 0
Work done to create the elongation ‘x’ is $W = F.x = (Kx)(x) = K{x^2}$
So this is converted to the energy and hence the final energy = $K{x^2}$
Now as the spring force varies linearly with elongation, force is not constant and it varies at every instant and it is the same in case of work done so work also varies which means energy at every instant also varies.
Even though energy at every instant is varying we know the initial energy and the final energy we take the average of to find out the energy stored.
In this wire case $x = l$ hence the final energy is $K{l^2}$
So average energy is $\dfrac{{0 + K{l^2}}}{2} = \dfrac{1}{2}K{l^2}$
So the work done in increasing the length is $\dfrac{1}{2}K{l^2}$

So, the correct answer is “Option C”.

Note: The energy stored in the wire is only half the work done. One might doubt where the other half of work has gone. The other half of the work is lost as heat. Same with the case of spring i.e half work is stored as internal potential energy and the remaining half is lost as heat.