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allsm [11]
3 years ago
13

A force of 18 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching i

t from its natural length to 11 in. beyond its natural length? W = ft-lb
Physics
1 answer:
bagirrra123 [75]3 years ago
8 0

Answer:

11.54 ft - lb

Explanation:

F = 18 lb, x = 8 in = 8 / 12 = 0.66 ft

F = k x

k = F / x = 18 / 0.66 = 27.27 lb/ft

y = 11 in = 11 / 12 ft = 0.92 ft

Work done = 1 /2 x k x  y^2

W = 0.5 x 27.27 x 0.92 x 0.92 = 11.54 ft - lb

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