Suppose that 2 j of work is needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. (a) how much work is needed to stretch the spring from 35 cm to 40 cm? (b) how far beyond its natural length will a force of 30 n keep the spring stretched?

(a) 1.042 J

b) 0.108 m

Explanation:

(a)

W = work done to stretch the spring = 2 J

k = spring constant

L₀ = initial length = 30 cm = 0.30 m

L = final length = 42 cm = 0.42 m

x = stretch of the spring = L - L₀ = 0.42 - 0.30 = 0.12 m

Work done to stretch the spring is given as

W = (0.5) k x²

2 = (0.5) k (0.12)²

k = 277.78 N/m

x₀ = initial stretch of the spring = 35 - 30 = 5 cm = 0.05 m

x = final stretch of the spring = 40 - 30 = 10 cm = 0.10 m

W' = work needed to stretch the spring from 35 cm to 40 cm

Work needed to stretch the spring from 35 cm to 40 cm is given as

W' = (0.5) k (x² - x₀²)

W' = (0.5) (277.78) (0.10² - 0.05²)

W' = 1.042 J

b)

x = Stretch of the spring beyond natural length

F = force = 30 N

Spring force is given as

F = k x

30 = (277.78) x

x = 0.108 m

force required to detach it is given by

charges are taken to be fundamental charge

now we have

so above is the minimum distance at which it will detach