Question

K is the force constant of a spring. The work done in increasing its extension from $l_{1}\ to\ l_{2}$ will be
(a) $K(l_{2}-l_{1})$
(b) $\frac{K}{2}(l_{2}+l_{1})$
(c) $K(l^{2}_{2}-l^{2}_{1})$
(d) $\frac{K}{2}(l^{2}_{2}-l^{2}_{1})$

Answer 1

Answer:

D) K/2 (l²2-l²1)

Explanation:

Force constant = K (Given)

Natural length = L (Given)

According to the work Energy Theorem.

Work Done = Final Total Energy – Initial Total Energy.

Where,

l1 = First extension.

l2 = Second extension.

Difference in first and second extension = l1 - l0.

Work done = (Kl1²)/2 – (Kl2²)/2.

On stretching, the spring will gain energy, thus the work done is the increase the energy.  Work done = Change in energy or Increase in energy.

Thus, the work done in increasing its extension from l1 to l2 is K/2 (l²2-l²1)