force of 10 lb is required to hold a spring stretched 4 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?

Given force=10lb

L1=4in converting to feet

But 0.08333ft = 1 inch

Then 4 inch is 0.3332

6inch is 0.49998

But hookes law states

F=Kx where F is force, K is the force constant, X

K=F/X=10/0.3333=30N/m

Integrating this

Integral of 30x with limit 0.333 to 0.5

F=30x^2/2=15x^2substing the limit

F = (15 (0.5^2-0.33^2) = 2.08ft-lb