Question

A wire fixed at the upper end stretches by length l by applying a force F. The work done in stretching is(a) $2Fl$(b) $Fl$(c) $\frac{F}{2l}$(d) $\frac{Fl}{2}$

Answer 1

Answer:

Fℓ/2

Explanation:

Length of the wire = l (Given)

Force applied = F (Given)

Let area of the wire = A

Thus, as per Young's modulus = Y = FL/Al

Therefore F = YAl/L

Work done by constant force in displacing the object by a distance ℓ  will be

= 1/2 ×stress × strain ×volume

= 1/2 / F/A ×ℓ/L ×A ×L

= Fℓ/2

Thus, the work done in stretching is  Fℓ/2

Answer 2

Answer:

Work done in stretching is Fl / 2 J

Explanation:

Given,

A wire fixed at the upper end stretches by length l by applying a force F.

To find: The work done in stretching is

Solution:

The work done in stretching is equal to half of the product of stretching force and the increase in length.

or, Work done = (1/2) * stretching force * increase in length

or, Work done = (1/2) * F * l

or, Work done = Fl / 2

Therefore, Work done in stretching is Fl / 2 J