Question

A force of 12 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 11 in. beyond its natural length?

Answer 1

Answer:

Work done in stretching the spring = 7.56 lb-ft.

Step-by-step explanation:

Normal length of the spring = 8 in or $\frac{8}{12}$ ft

= $\frac{2}{3}$ ft

If the spring has been stretched to 11 inch then the stretched length of the spring is = 11 in

= $\frac{11}{12}$ ft

Force applied to stretch the spring = 12 lb

By Hook's law,

F = kx [where k is the spring constant and x = length by which the spring is stretched]

12 = k($\frac{2}{3}$)

k = 18

Work done (W) to stretch the spring by $\frac{11}{12}$ ft will be

W = $\int\limits^\frac{11}{12} _0 {kx} \, dx$

    = $\int\limits^\frac{11}{12} _0 {(18x)} \, dx$

    = $18[\frac{x^{2}}{2}]^{\frac{11}{12}}_0$

    = 9($\frac{11}{12}$)²

    = 7.56 lb-ft