A force of 128 lb is required to hold a spring stretched 2 ft beyond its natural length. How much work W is done in stretching i

Question

2 years ago

8

t from its natural length to 9 inches beyond its natural length

1 answer:

Agata [3.3K] · Answer 1 · 2023-05-28T10:36:47+03:00

We have that the workdone in stretching natural length to is mathematically given as

W=9/2lbft

<h3>Workdone in stretching natural length</h3>

Question Parameters:

A force of 128 lb is required to hold a spring stretched 2 ft <em>beyond </em>its natural length.
Stretching it from its natural length to 9 inches beyond its natural length

Generally the Hookes equation for the Force is mathematically given as

F=Kx

Where

3.2lb=2ft*k

Thereofore

k=16

Force becomes

f=kx

f=16x

Hence

$W=8[x^2]^{3/4}_{0}$

W=9/2lbft

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