Elastic vs Gravitational Energy of a System

[GoGoGadget]

Homework Statement

I had a question regarding the elastic potential energy of a spring to the gravitational force of a cart. If you had a spring that was attached to the top of the inclined plane and to a cart with the cart moving downhill on an incline that was at an angle, how would finding the maximum extension of the spring and maximum velocity help in determining whether elastic and gravitational energies were equivalent or not? I know that if you were to take the cart at one position being at the top of the incline, potential energy would be zero, as would would Einitial. Efinal would then be equivalent to 1/2mv^2. But then when you have the cart (and spring) at the bottom of the incline, you could show potential energy with the cart, it being mgx. However, I'm not sure how this would allow you to help determine whether or not mgx is equivalent to 1/2kx^2 (the spring?)? Or what finding max velocity would tell you? Any input would be wonderful!

 

[rude man]

At points along the incline as the cart wheels on down that incline, the speed picks up and the spring starts to stretch. All along the incline, potential + kinetic + spring energy = mgh = constant. So for example at the top before the cart is released, p.e. = max. and k.e. and spring-stored energy = 0. Then, at the moment when the cart stops moving near the bottom (h = 0), all the p.e. lost (= mgh) is stored in the spring. At intermediate points there is finite p.e., k.e. and spring energy. But always the sum = mgh.

At the bottom (v = 0) you could measure spring extension x and height loss h, then verify by calculation that 1/2 kx^2 = mgh. Remember, this is when the cart stops moving downhill (v = 0) before lurching back up. Of course, you ignored friction losses so your computation will not come out exact.