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

Work Done by a Spring Force — Question 7

Section 7-6 · Physics pages 964–965

Question 7 Parallel springs contribute separate amounts of work Assume ideal massless Hookean springs with fixed supports unless stated otherwise. Two horizontal springs with stiffnesses 100 and 300 N/m connect the same fixed wall to a rigid 2 kg slider. Both are relaxed at x=0. Released from rest at x=0.10 m, the slider moves without friction. 2 kg 300 N/m 100 N/m Parallel springs share the same slider displacement. (a) Find work by each spring before the slider reaches x=0. (b) Find the slider speed there. (c) Find the effective stiffness and explain why averaging the two stiffnesses is wrong.
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Question 7 - Solution Parallel springs contribute separate amounts of work 2 kg 300 N/m 100 N/m (a) Use each spring deformation. The governing relation is W1 + W2 = 1 2(k1 + k2)(x2 i −x2 f) W1 = 1 2(100)(.10)2 = 0.50 J, W2 = 1 2(300)(.10)2 = 1.50 J. (b) Sum all work on the common slider. 1 2(2)v2 = 2 J, so v = √ 2 m/s. (c) Sum forces for the same displacement. F = −(k1 + k2)x = −400x, so keff= 400 N/m. Both springs act simultaneously through the full displacement; no division by two is warranted.

Question 7

Parallel springs contribute separate amounts of work

Assume ideal massless Hookean springs with fixed supports unless stated otherwise. Two horizontal springs with stiffnesses 100 and 300 N/m connect the same fixed wall to a rigid 2 kg slider. Both are relaxed at x=0. Released from rest at x=0.10 m, the slider moves without friction.

Diagram shown in the original SVG worksheet.

  1. Find work by each spring before the slider reaches x=0.

  2. Find the slider speed there.

  3. Find the effective stiffness and explain why averaging the two stiffnesses is wrong.

Question 7 - Solution

Parallel springs contribute separate amounts of work

Diagram shown in the original SVG worksheet.

(a) Use each spring deformation.

The governing relation is W1+W2=12(k1+k2)(xi2−xf2)\boxed{W_1+W_2=\tfrac12(k_1+k_2)(x_i^2-x_f^2)}

W1=12(100)(.10)2=0.50J,W2=12(300)(.10)2=1.50J.W_1=\tfrac12(100)(.10)^2=0.50\ \mathrm J,\qquad W_2=\tfrac12(300)(.10)^2=1.50\ \mathrm J.

(b) Sum all work on the common slider.

12(2)v2=2J,\tfrac12(2)v^2=2\ \mathrm J, so v=2m/s.v=\sqrt2\ \mathrm{m/s}.

(c) Sum forces for the same displacement.

F=−(k1+k2)x=−400x,F=-(k_1+k_2)x=-400x, so keff=400N/m.k_{\mathrm{eff}}=400\ \mathrm{N/m}. Both springs act simultaneously through the full displacement; no division by two is warranted.

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