Arun
Last Activity: 7 Years ago
Dear Harsh
Assuming Hookes law is obeyed throughout the stretch, the force required to stretch the spring is F = kx where k is the spring constant and x the amount of stretch.
If you imagine plotting a graph of force against x, the work done will be the integral of the graph dx with limits from 0.05 m to 0.1 m (the unit of work and spring constant involves metres so you have to convert these), since Hookes law is obeyed, the graph is linear and so the integral is simply 1/2 *kx^2 so work done is [1/2 *kx^2] upper limit 10 and lower limit 5 = (1/2 *k*0.1^2)-(1/2 *k*0.05^2) = 18.75 Nm
Regards
Arun (askIITians forum expert)
