Question

An object with a mass of `7 kg` is hanging from a spring with a constant of `2 (kg)/s^2`. If the spring is stretched by `1 m`, what is the net force on the object?

Answer 1

When the spring is stretched by `x`, the amount of restoring force (`F`) acting against the direction of displacement (here,upwards) is `K*x=2*1=2 N` (where, `K` is the spring constant)

Now,as the object is attached to the spring,so on stretching the spring by `1m` its weight will be opposed by the restoring force generated in the spring.

So,net force acting will be `(7*10 - 2)=68N` downwards