The potential energy of a spring when stretched through a distance x is 10J. What is the amount of work done on the same spring to stretch it through on additional distance x?

Explanation

P.E of the spring when stretched through a distance x,
`u=(1)/(2)kx^(2)=10J`
When x becomes 2x, the potential energy will be
`u'=(1)/(2)k(2x)^(2)=4xx(1)/(2)kx^(2)=4xx10`
= 40J
`therefore` Work done = `u'-u=40-10=30J`