Suppose that 2 J of work are needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. How far (in cm) beyond its natural length will a force of 30 N keep this spring stretched

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched

Explanation:

Work, W = 2 J

Initial distance, = 30 cm

Final distance,  = 42 cm

Force, F = 30 N

Stretched length, x = ?

We know,

W = 1/2 kΔx²

Δx = 42-30 cm = 12 cm = 0.12 m

2 J = 1/2 k X (0.12)²

k = 277.77 N/m

According to Hooke's law,

F = kx

30 N = 277.77 X x

x = 0.108 m

x = 10.8  cm

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.

= initial pressure

= final pressure = 2

= initial volume

= final volume = /4

= initial temperature = 20 c = 20 + 273 = 293 k

= final temperature = ?

using the equation