Answer:
PE = 11.25 J
Explanation:
Elastic Potential Energy
It's the energy stored in an elastic material like a spring of constant k, in which case the energy is proportional to the square of the change of length Δx and the constant k.
[tex]\displaystyle PE = \frac{1}{2}k(\Delta x)^2[/tex]
The car spring has a constant of k=90 N/m and it stretches Δx=0.5 m. The elastic potential energy is:
[tex]\displaystyle PE = \frac{1}{2}\ 90(0.5)^2[/tex]
Calculating:
PE = 11.25 J
Elastic potential energy is the potential energy held when an elastic item is stretched or compressed by an external force. The elastic potential energy of a car spring will be 11.25 J.
Elastic potential energy is the potential energy held when an elastic item is stretched or compressed by an external force. It is equal to the work done to extend the spring.
The given data in the problem is;
x is the distance by which spring got stretched = 0.5 m
k is the spring constant for the car = 90 N/m.
The formula for the elastic potential energy is;
[tex]\rm PE =\frac{1}{2} kx^{2} \\\\ \rm PE =\frac{1}{2} \times 90 (0.5)^{2} \\\\ \rm PE = 11.25\; J[/tex]
Hence the elastic potential energy of a car spring will be 11.25 J.
To learn more about the elastic potential energy refer to the link;
https://brainly.com/question/156316
