A force of 3 pounds is required to hold a spring stretched 0.6 feet beyond its natural length. how much work (in foot-pounds) is

Question

1 year ago

7

done in stretching the spring from its natural length to 0.7 feet beyond its natural length? answer exactly or round to 2 decimal places.

1 answer:

ioda · Answer 1 · 2024-01-30T01:49:44+03:00

The work done (in foot-pounds) in stretching the spring from its natural length to 0.7 feet beyond its natural length is 1.23 foot-pound

<h3>Data obtained from the question</h3>

From the question given above, the following data were obtained:

<h3>How to determine the spring constant</h3>

F = Ke

Divide both sides by e

K = F/ e

K = 3 / 0.6

K = 5 pound/foot

Thus, the spring constant of the spring is 5 pound/foot

<h3>How to determine the work done</h3>

Wd = ½Ke²

Wd = ½ × 5 × 0.7²

Wd = 2.5 × 0.49

Wd = 1.23 foot-pound

Therefore, the work done in stretching the spring 0.7 feet is 1.23 foot-pound

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