# Stretching a spring

A force of 52N stretches a spring 0.73m from equalibrium. What is the value of the spring constant?

F=-k * delta s
52N= -k * 0.73m
52N/0.73N = -k
k= -71.2 N/m or -71 N/m with sig figs.
The negative sign means that the force is opposite of the stretch.

Is this right?

Dick
Homework Helper
Of course it's right.

alphysicist
Homework Helper
The spring constant k will be a positive number.

The force F in Hooke's law is the force of the spring, not the external force that pulls on the spring. Also, F and (delta s) are vectors so we need to keep track of their directions.

So, just to pick a specific direction, let the spring be pulled to the right, and also choose rightwards to be the positive direction. Then (delta s)=+0.73m. But to get the sign of F we need to find which direction the spring is pulling.

So:
both the force and the displacement are in the same direction since the sping is being stretched, but the k constant is pulling in the opposite direction so k = 71 N/m pulling the spring in the opposite direction of the stretch.

Last edited:
alphysicist
Homework Helper
If you pull a spring to the right, the spring pulls back against you to the left. So if delta s is positive, then F is negative.

ok..

-52N=-k*0.73m
-52N/0.73m = -k
-71 N/m = -k
71 N/m = k

Which is what I got before.

alphysicist
Homework Helper
If my posts made it seem like I was saying the magnitude of your answer was wrong, I definitely did not intend that. I was just pointing out that the sign of your original answer was wrong, and that it might be coming about because maybe you were thinking that F was the external, applied force when it is really the force of the spring.