Calculating Spring Constant from 52N Force and 0.73m Stretch

[StephenDoty]

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]

Of course it's right.

 

[alphysicist]

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.

 

[StephenDoty]

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.

 

[alphysicist]

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.

 

[StephenDoty]

ok..

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

Which is what I got before.

 

[alphysicist]

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.