Question

A spring with a 3.15kg weight hanging from it measures 13.40cm, and without the weight 12.00cm. If you hang a weight on it so as to store 10.0J potential energy in it, how long will the spring be?

Answer 1

Final Length = 12.45 cm

Step-by-step explanation:

First we need to find the spring constant. From Hooke's Law:

F = kΔx

where,

F = Force Applied on Spring = Weight = mg = (3.15 kg)(9.8 m/s²) = 30.87 N

k = spring constant = ?

Δx = change in length of spring = 13.4 cm - 12 cm = 1.4 cm = 0.014 m

Therefore,

30.87 N = k(0.014 m)

k = (30.87 N)/(0.014 m)

k = 2205 N/m

Now, for the Potential Energy of 10 J:

P.E = (1/2)KΔx²

where,

P.E = Potential Energy of Spring = 10 J

Δx = ?

Therefore,

10 J = (2205 N/m)Δx²

Δx = √[10 J/(2205 N/m)

Δx = Final Length - Initial length = 0.0045 m = 0.45 cm

Final Length = 0.45 cm + 12 cm

Final Length = 12.45 cm

Answer 2

21.52 cm

Step-by-step explanation:

Given that

mass of the spring, m = 3.15 kg

Length of the spring l2, = 13.4 cm = 0.134 m

Length of the spring l1 = 12 cm = 0.12 m

change in extension, x = 0.134 - 0.12 = 0.014 m

Acceleration due to gravity, g = 9.8 m/s²

Potential Energy, U = 10 J

See attachment for calculation