Question

A block weighs 15 n and is suspended from a spring that is attached to the ceiling. the spring stretches by 0.075 m from its unstrained length. by how much does the spring stretch when a 33-n block is suspended from it?

Answer 1

We can salve the problem by using the formula:

`F=kx`

where F is the force applied, k is the spring constant and x is the stretching of the spring.

From the first situation we can calculate the spring constant, which is given by the ratio between the force applied and the stretching of the spring:

`k=(F)/(x)=(15 N)/(0.075 m)=200 N/m`

By using the value of the spring constant we calculated in the first step, we can calculate the new stretching of the spring when a force of 33 N is applied:

`x=(F)/(k)=(33 N)/(200 N/m)=0.165 m`