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

Explanation:

answer :

b

explanation:

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