On stretching a spring by a distance L it's potential energy increases by 10J.The energy required to further stretches it by a distance l,will be?

1 Answer
Feb 10, 2018

If a spring of spring constant (#K#) is stretched by a length of #L#,the elastic potential energy stored in it #E# is given as, #E=1/2 KL^2#

Given, #E=10J#,

so, #10=1/2 KL^2#

so, #K= 20/L^2#

So,to stretch it further by a length of #l# if energy required is #E"#,then, #E"= 1/2 Kl^2= 1/2 (20/L^2) l^2=10 (l/L)^2#