Question

A spring is characterized by a spring constant of 60 N/m. How much potential energy does it store, when stretched by 1.0 cm?

Answer 1

Answer:

3.0 x10^-3 J

Explanation:

The potential energy of a spring is given by PE = (0.5)k*x^2  

Where

K: Spring Constant = 60 N/m

x: displacement of the spring from its equilibrium position = 1cm = 0.01m

Then PE = 0.5(60)(.01)^2 = 0.003J = 3.0 x10^-3 J

Answer 2

Answer:

The potential Energy stored in the spring = 0.003 J

Explanation:

Potential Energy: This is the energy possessed by a body by virtue of its position. The ability to stretched or compressed an elastic material to do work is called the elastic potential energy. The unit of elastic potential energy is Joules (J).

Elastic potential energy can be represented mathematically as

W = 1/2ke² ............................... Equation 1

Where W = Elastic potential Energy, k = force constant of the spring. e = extension.

Given: k = 60 N/m, e = 1.0 cm = (1/100) m = 0.01 m.

Substituting these values into equation 1,

W = 1/2(60)(0.01)²

W = 30×0.0001

W = 0.003 J.

Thus potential Energy stored in the spring = 0.003 J