Physics Dynamics review

Question

A steel spring is stretched from length 3.0 cm to a length of 8.0 cm. If the spring constant is 200 N/m, what force is required? Make sure you draw a FBD and write Fnet = ma to begin this solution.

Answer 1

Hi Jamie!

 

I am not sure what they mean by the "force" required.  Because the spring pulls back with an increasing force as it is stretched more, it takes a different force at every interval.  The only way I know that talking about the force would make sense is if they were asking for the force to hold it at 8.0 cm.

 

Otherwise, I would think they would ask for the work needed to stretch this.

 

If they are asking for some total force over all the stretching, then that would also come out to be the work needed, which equals change in spring potential energy.

 

But I would not see much use in starting with a free-body diagram or setting up Newton's second law in this situation, especially if no specific acceleration is asked for in the stretching.

 

In any case, the work needed to stretch it as described (which would also be the integral of the force over every little differential bit "dx" of stretching) is equal to the change in potential energy of the spring between 0.03 m and 0.08 m.

 

W = ΔPE = (1/2)kx_f² - (1/2)kx_o² = (1/2)(200 N/m)((0.08 m)² - (0.03 m)²) = 0.55 J

 

I hope that helps some.  I will think about this more, and perhaps another tutor will have different insight, as well.