Aforce 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

m∠5 + m∠3 = m∠1   is true.

m∠5 + m∠6 = 180 is true

m∠2 +m∠3 = m∠6   is true

m∠2 +m∠3 +m∠5 = 180   is true

thats one

step-by-step explanation: