Question

A spring of force constant 1500 Nm-l is acted

Answer 1

Answer:

1.876 J

Explanation:

First, let’s calculate the compression of the spring from the Hooke’s law:

F=kx,

here, F=75 N is the force acted on the spring, k=1500 N⁄m is the force constant of the spring, x is the compression of the spring.

Then, we get:

x=F/k=(75 N)/(1500 N/m)=0.05 m.

Finally, we can find the potential energy stored in the spring:

PE=1/2 kx^2=1/2∙1500 N/m∙(0.05 m)^2=1.875 J.

correct my answer if it's wrong ^^