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

`U=(1)/(2)kx^(2)=10J`
When x becomes 2x, the P.E will be,
`U'=(1)/(2)k(2x)^(2)=4((1)/(2)kx^(2))`
`=4xx10=40J`
`therefore` Work done = `U'-U`
`=40-10=30J`