# Net Force vs Net Work

I hope you are doing well :)

When I was practicing, I came upon this question:

On a horizontal frictionless surface there is a Spring with a Spring constant is 50 N/m. Initially, the spring is at its relaxed length and a block is stationary at position x = 0. Then an applied force with a constant magnitude of 3.0 N pulls the block in the positive direction of the x-axis, stretching the spring until the block stops. When that stopping point is reached, what is the position of the block?

The question is similar to to this post: Why equating forces give wrong answer?, except that the question in the other post asked for the maximum displacement, and since the system is dynamic, may not be when the block comes to rest. However, the question in this post directly indicates that the block comes to rest.

I thought that since the block was stationary, the applied force would cancel out with the spring force, leading to zero acceleration, so Fapplied + Fspring = 0 ... so doing that, I got -kx + 3 = 0 and since k = 50, -50x + 3 = 0, so x = 3/50 = 0.06 meters. However, the answer key showed that the Work cancels out, meaning Wapplied + Wspring = 0, and got x = 0.12 meters. I would like to know why the Forces don't cancel out, in this scenario, if the block is literally at rest? It would be awesome if someone could help clear my doubt, at your convenience. Thanks, and have a great day!

• How did calculate $W_{applied} + W_{spring} = 0$?
– Gert
Commented Jan 15, 2021 at 22:11
• The question is very poorly worded. Commented Jan 16, 2021 at 0:57
• Hi! Because the block is initially stationary (at x = 0), and the question indicates that the block becomes stationary again, this means that, at that point, Wapplied + Wspring = 0. The question asks what the block's position is when the block is stationary for the 2nd time, meaning when Wapplied + Wspring = 0. Thanks! Commented Jan 16, 2021 at 17:16

The equation of motion is $$m \ddot x = F - kx$$ where $$m$$ is mass, x is the extension of the spring from its relaxed position and $$F$$ is the applied force. Starting from rest at $$x = 0$$, the solution is $$x(t) = {{F(1 - cos( {\sqrt{k \over m}})t)} \over k}$$. This is oscillatory motion. $$x(t)$$ is maximum at each t where $${\sqrt{k \over m}t} = n \pi$$; that is, for every $$t = {\sqrt{m \over k}n \pi}$$ where n is an odd integer 1, 3,5 ... At each of these times $$x = {{2F \over k}} = {6 \over {50}} = 0.12$$. The maximum extension is not when the applied force and the spring restoring force are equal, because the mass is moving and has inertia. Notice how much simpler the solution is using energy instead of the equation of motion; however, both approaches yield the same results.