Question

A force of 18 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 10 in. beyond its natural length

Answer 1

Answer: 18.75 lb.ft

Step-by-step explanation:

Given

Force required to stretch spring 8 in. is 18 lb

it can be written

$\Rightarrow F=kx\\\Rightarrow 18=k(8)\\\\\Rightarrow k=\dfrac{18}{8}=\dfrac{9}{2}\ lb/in.$

Work done in stretching from its natural length to 10 in.

$\Rightarrow W=\dfrac{1}{2}kx^2\\\\\Rightarrow W=0.5\times \dfrac{9}{2}\times (10)^2\\\\\Rightarrow W=225\ lb.in.\ or\\\Rightarrow W=18.75\ lb.ft$