Question

A force of 10 lb is required to hold a spring stretched 2 in. beyond its natural length. How much work W is done in stretching it from its natural length to 5 in. beyond its natural length?

Answer 1

Work done will be equal to 5.2059 lb-ft

Explanation:

We have given force F = 10 lb

Spring is stretched to 2 in

So x = 2 in

As 1 inch = 0.0833 feet

So 2 inch = 2×0.0833 = 0.1666 feet

From hook's law we know that F = Kx , here K is spring constant and x is spring elongation

So
`10=K* 0.1666`

K = 60.024 lb/feet

Now new elongation x = 5 in

So 5 in = 5×0.0833 = 0.4165 feet

Work done is given by
`W=(1)/(2)Kx^2`

So
`W=(1)/(2)* 60.02* 0.4165^2=5.205lb-ft`

So work done will be equal to 5.2059 lb-ft