Question

Please HELP! WILL GIVE BRAINLIEST!
A mass of 500 g is attached to a spring with elastic constant coefficient of elasticity = 50.0 N / m. The spring is lengthened by 10 cm and then left free. Calculate the return force exerted by the spring.​

Answer 1

Answer:

Force, F = -5 N                  

Explanation:

We have,

Mass, m = 500 g = 0.5 kg

Spring constant of the spring, K = 50 N?M

The spring is lengthened by 10 cm and then left free.

It is required to find the return force exerted by the spring. It is based on Hooke's law. The force on the spring is given by :

$F=-kx\\\\F=-50\times 0.1\\\\F=-5\ N$

So, the return force exerted by the spring is (-5 N)