Understanding Work in Physics: Misconceptions and Correct Approaches

[GarrettB]

Hi,

So I was able to answer both of these questions eventually, but the second one gave me some trouble. At first, I tried finding the force that the spring would be applying at that stretched length (3.81cm), knowing that the force done by external agent must be equal and opposite. I then tried multiplying the applied force by the distance it would do work across in order to find the amount of work. Unfortunately this did not work and instead realized I could find the work done by the spring (1/2kx^2) directly instead. I'm wondering, conceptually, why is the first approach wrong? Thanks a lot.

When a 3.89kg mass is hung vertically on a certain light spring that obeys Hooke's law, the spring stretches 2.87cm. If the 3.89kg mass is removed how far will the spring stretch if a 1.48kg is hung on it instead? 1.09×10-2 m

You are correct. Previous Tries
How much work must an external agent do to stretch the spring 3.81cm from its unstretched position? 9.65×10-1 J

 

[Haborix]

So imagine you start the spring at equilibrium. Now, slowly start to stretch the spring, is the force the same or is it changing as you move it away from equilibrium? Your technique of force times distance will work (no pun intended), but you have to do a little integration. You should do this integration of the force along the displacement and I think you'll be pleasantly surprised by the result.

 

[GarrettB]

Oh right, the force is always changing. Thanks for the insight!