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

5.5 lb-ft

Step-by-step explanation:

k = Spring constant

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

F = Force = 11 lb

`F=kx\\\Rightarrow k=(F)/(x)\\\Rightarrow k=(11)/((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[(44x^2)/(2)\right]^(0.5)_0\\\Rightarrow W=(44* 0.5^2)/(2)=5.5`

Work done is 5.5 lb-ft