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Physics / Chapter 7: Kinetic Energy and Work / Section 7-6

Work Done by a Spring Force — Question 8

Section 7-6 · Physics pages 966–967

Question 8 An unattached spring stops pushing at its relaxed length A 0.50 kg block rests against an unattached horizontal spring compressed by 0.10 m. The spring has k = 200 N/m and can push but cannot pull the block. Release occurs from rest on a frictionless rail. m x The block is in pushing contact only; it is not fastened to the spring. (a) Find the speed when contact is lost. (b) Find speed 0.30 m beyond the relaxed position. (c) Explain why extending F = −kx into positive x would predict an unphysical return.
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Question 8 - Solution An unattached spring stops pushing at its relaxed length m x (a) Use work over the contact interval. The governing relation is Fs(x) = { −kx, x ≤0 during contact, 0, x > 0 after separation. Ws = 1 2(200)(.10)2 = 1 J. Thus 1 2(.50)v2 = 1, giving v = 2 m/s at x=0. (b) Change the force law after separation. Beyond x=0 the spring force on the block is zero. No horizontal force acts, so speed remains 2 m/s. (c) Keep contact and attachment distinct. A continuation F = −kx for x>0 assumes a tensile attachment that does not exist. The actual force is −kx while compressed and zero after the block departs; there is no spring-driven return.

Question 8

An unattached spring stops pushing at its relaxed length

A 0.50 kg block rests against an unattached horizontal spring compressed by 0.10 m. The spring has k=200k=200 N/m and can push but cannot pull the block. Release occurs from rest on a frictionless rail.

Diagram shown in the original SVG worksheet.

  1. Find the speed when contact is lost.

  2. Find speed 0.30 m beyond the relaxed position.

  3. Explain why extending F=−kxF=-kx into positive x would predict an unphysical return.

Question 8 - Solution

An unattached spring stops pushing at its relaxed length

Diagram shown in the original SVG worksheet.

(a) Use work over the contact interval.

The governing relation is Fs(x)={−kx,x≤0during contact,0,x>0after separation.\boxed{F_s(x)=\begin{cases}-kx,&x\leq0\ \text{during contact},\\0,&x>0\ \text{after separation}.\end{cases}}

Ws=12(200)(.10)2=1J.W_s=\tfrac12(200)(.10)^2=1\ \mathrm J. Thus 12(.50)v2=1,\tfrac12(.50)v^2=1, giving v=2m/sv=2\ \mathrm{m/s} at x=0.

(b) Change the force law after separation.

Beyond x=0 the spring force on the block is zero. No horizontal force acts, so speed remains 2 m/s.

(c) Keep contact and attachment distinct.

A continuation F=−kxF=-kx for x>0 assumes a tensile attachment that does not exist. The actual force is −kx-kx while compressed and zero after the block departs; there is no spring-driven return.

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