Question

A force of 8 lb is required to hold a spring stretched 8 in. beyond its natural length. How much work W is done in stretching it from its natural length to 14 in. beyond its natural length?

Answer 1

work done=-8.166lb-ft

Step-by-step explanation:

force = 8 lb spring stretched (x)= 8 in

according to hooks law

F=-kx

k=-
`(f)/(x)`

k=-
`(8)/(8)`

k=-1 lb/in

work done=
`\int_(0)^(14)kx\ dx`

work done=
`(-1)\left [(x^2)/(2)  \right ]^(14) _(0)`

work done=-98lb-in

1 ft =12 in

work done=
`(-98)/(12)`

=-8.166lb-ft