Question

If the potential energy of a spring when stretched through a distance ‘a’ is 25 J, then
what is the amount of work done on the same spring so as to stretch it by an additional
distance ‘5a’?

Answer 1

P.E =1/2 (spring constant )(distance from equilibrium )²

U=1/2 K x²
where U=P.E of spring
k=spring constant
x=distance the spring is stretched.

According to given,
U=25J
1/2ka²=25J

To stretch through additional distance 5a
=(1/2 )k x36a²
=36 (1/2 )k a²
=36x25=900J

Additional work done=900-25=875J