Question

when a load of 10kg is hung from the wire then extension of 2m is produced the work done by the restoring force

Answer 1

we know, restoring force is given by hook's law.
e.g., restoring force , F = kx
and weight = mg
here restoring force is balanced by weight
so, kx = mg
k = (mg/x)

here m = 2kg , g = 10m/s² and x = 2m
so, k = (10 × 10/2) = 50N/m

now , workdone by restoring force = 1/2 kx²
= 1/2 × 50 × (2)²
= 1/2 × 50 × 4
= 100J

Answer 2

Given:

Load = 10 kg

Extension of 2 m

To find:

The work done by restoring force

Solution:

By Hook's law,

F = k . Extension length

Where,

F - restoring force

Weight = mg

Where,

m - mass

g - gravity

The restoring force is cancelled by weight.

They can be replaced.

And so,

k . extension length = mg

k = (mg/ extension length)

Substituting the values,

k = (10 × 10/2)

k =50 N / m

To find the work done by restoring force,

W = 1/2 k . extension length^2

W = 1/2 × 50 × (2)^2

W = 1/2 × 50 × 4

W = 100 J

Hence, when a load of 10 kg is hung from the wire then extension of 2 m is produced the work done by the restoring force is 100 J.