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
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