A force of 10 lb is required to hold a spring stretched 4 in. beyond its natural length. How much work is done in stretching it from its natural length to 6 in. beyond its natural length?

W = 15/4 ft. ib

Step-by-step explanation:

Force = 10ib

According to hooked law, f (x) = kx

x = 4inches = 4/12 ft

x = 1/3ft

f (x) = 1/3k

10 = 1/3k

k = 30 ib/ft

f (x) = 30x

Workdone = integral of f (x) with its limit

6 inches = 6/12 ft

= 1/2ft

W = integral (1/2 to 0) of 30x

W = 15x^2 (1/2 to 0)

W = 15 (1/2) ^2 - 15 (0) ^2

W = 15 (1/4) - 0

W = 15/4 ft. Ib