Work Problem
2009-09-13 10:08 am
A force of 16 lb is required to hold a spring stretched 2 in. beyond its natural length. How much work W is done in stretching it from its natural length to 6 in. beyond its natural length?
回答 (1)
2009-09-13 4:39 pm
✔ 最佳答案
1 lb = 4.448 N1 in = 0.0254 m
Firstly, by Hooke's law,
F = kx
16 X 4.448 = k(2 X 0.0254)
Spring constant, k = 1.4 X 103 Nm-1
Now, the required work done
= Elastic potential energy gained
= 1/2 kx2
= 1/2 (1.4 X 103)(6 X 0.0254)2
= 16.3 J
參考: Physics king
收錄日期: 2021-04-19 15:18:59
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