Question

A force of 11 lb is required to hold a spring stretched 3 in. beyond its natural length. How much work W is done in stretching it from its natural length to 6 in. beyond its natural length?

Answer 1

Answer:

5.5 lb-ft

Explanation:

k = Spring constant

x = Stretched length = 3 in = 3/12 = 1/4 feet

F = Force = 11 lb

$F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{11}{\frac{1}{4}}\\\Rightarrow k=44$

So,

$F=44x$

Work done when x = 6 in = 6/12 = 0.5 feet

$W=\int_0^6Fdx\\\Rightarrow W=\int_0^{0.5} 44xdx\\\Rightarrow W=\left[\frac{44x^2}{2}\right]^{0.5}_0\\\Rightarrow W=\frac{44\times 0.5^2}{2}=5.5$

Work done is 5.5 lb-ft