Question

If the work required to stretch a spring 2 ft beyond its natural length is 12 ft-lb, how much work is needed to stretch it 9 in. beyond its natural length?

Answer 1

Answer:

Work done = 1.69 lb-ft.

Step-by-step explanation:

Work required to stretch a spring 2 feet beyond its natural length W = 12 ft-lb

By Hook's law

$W=\int\limits^2_0 {kx} \, dx$

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

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

12 = 2k

k = 6 lb per feet

Now we have to find the work done to stretch the spring 9 inch or $\frac{9}{12}$ feet beyond its natural length.

We again use the Hook's law.

$W=\int\limits^\frac{3}{4} _0 {kx} \, dx$

$W=k\int\limits^\frac{3}{4} _0 {x} \, dx$

$W=6\int\limits^\frac{3}{4} _0 {x} \, dx$

$W=6[\frac{x^{2} }{2}]^{\frac{3}{4}}_{0}$

$W=3(\frac{3}{4})^{2}$

$W=\frac{27}{16}$ lb-ft

Therefore, work done to stretch the spring 9 in. beyond the its natural length will be 1.69 lb-feet.