Question

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 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.

Answer 1

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

Data obtained from the question

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

• Force (F) = 3 pounds
• Extension (e) = 0.6 feet
• Work done (Wd) =?

How to determine the spring constant

• Force (F) = 3 pounds
• Extension (e) = 0.6 feet
• Spring constant (K) =?

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

How to determine the work done

• Spring constant (K) = 5 pound/foot
• Extention (e) = 0.7 feet
• Work done (Wd) =?

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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