The block in the figure below lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 60 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.7 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops.

Question

The block in the figure below lies on a horizontal frictionless surface and is attached to the free end of the spring, with a spring constant of 60 N/m. Initially, the spring is at its relaxed length and the block is stationary at position x = 0. Then an applied force with a constant magnitude of 2.7 N pulls the block in the positive direction of the x axis, stretching the spring until the block stops.
Assume that the stopping point is reached.

(a) What is the position of the block?
(b) What is the work that has been done on the block by the applied force?
(c) What is the work that has been done on the block by the spring force?

Im doing this online and I got these answers:
a.) .045m
b.) 0.122 J
c.) -0.06075 J

The website says they are wrong but I have absolutely no clue why, especiall for part A, its a joke. All you need to do is plug in F = -kx and they give you force and k but it still says its wrong.

michael · Answer

you need to take the integral of the force to find the work and then solve for x. &int -kx = &int 2.7 (-kx²) = 2.7x solve for x

michael · Answer

sorry, didn't write that properly..

(kx²)/2 = 2.7x

you solve for work because the block started from rest and ended at rest. KE_f - KE_i = Work_a + (-W_s)

where Work_a = the Work done by the applied force