Question 8

[asfino12217]

8.) If k=40000N/m, then how much work does Substance Q42 do in compressing from its relaxed length to a length of 9 cm?

For this question, it shows that the formula W=1/2kx^2 was used but why couldn't this problem be calculated by figuring out what the force was and then multiplying it by distance (W=Fd).

[nedaa.asba6809]

We are actually looking at the potential energy stored in a spring, the spring constant K, is a measure of the stiffness of a spring. Therefore, in order to compress a spring by a distance Δx. We may apply a Force Fext = kΔx and the minus sign comes in from Newton’s 3rd law and because the force is always in the opposite direction from the displacement; that’s the linear restoring force.
Fext = ½ (Fo + Fx) = ½ Kx
PEs = ½ kx2
PEs = 0 when (x = 0) it is at equilibrium.
PEs > 0 [when the spring is not in its equilibrium].
PEs is the same if x = ± xf (same PEs for equal expansion or compression).

[admin]

Bump.