A force of 7 pounds is required to hold a spring stretched 0.4 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?

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jtellezd jtellezd · Answer 1 · 2019-11-23T13:50:38+01:00

Answer:

W = 8.575 foot-pounds

Step-by-step explanation:

The equation for work is: W = F*d

where F is force applied and d distance of the movement

from problem statement we have

d = 0,7 ft

We also know that the spring was stretched 0,4 ft when a force of 7 pounds was applied therefore

K the constant of the spring is

F = K*s

k = F/s ⇒ k = 7/0.4 ⇒ k = 17.5 pounds/feet

Now to move from original condition of the spring up to 0,7 feet we need a force of

F = k*s ⇒ F = 17,5 pounds/feet * 0.7 feet ⇒ F = 12.25 pounds

And finally the work

W = 12.25 * 0.7 = 8.575 foot-pounds

W = 8.575 foot-pounds