Suppose we have a spring with one end fixed and block of mass m is attached with the other free end. This position of the free end is x=0. The spring constant k=50 N/m.

Now if we apply a constant force of 100N on the block to displace it rightward, the spring stretches and applies leftward force on the block. By Hook’s statement we can find the displacement of the block from x=0 and that is x=2 (lets treat x=2 as meters, so 2 meter displacement). After two meter displacement the block stops (because forces are balanced now).

Now since the block is stationary before and after the displacement so we can apply the work energy theorem, By work energy theorem we get

Work done by Applied force= 200 J

Work done by Spring force = -200 J

Now if we apply the integral formula as integral of –kxdx from 0 to 2 we different result that is

Work done by spring force = -100 J

Why this is so? Can we not apply the integral formula here? Or what’s the problem?

Suppose we have a spring with one end fixed and block of mass m is attached with the other free end. This position of the free end is x=0. The spring constant k=50 N/m.

Now if we apply a constant force of 100N on the block to displace it rightward, the spring stretches and applies leftward force on the block. By Hook’s statement we can find the displacement of the block from x=0 and that is x=2 (lets treat x=2 as meters, so 2 meter displacement). After two meter displacement the block stops (because forces are balanced now).

Now since the block is stationary before and after the displacement so we can apply the work energy theorem, By work energy theorem we get

Work done by Applied force= 200 J

Work done by Spring force = -200 J

Now if we apply the integral formula as integral of –kxdx from 0 to 2 we different result that is

Work done by spring force = -100 J

Why this is so? Can we not apply the integral formula here? Or what’s the problem?