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wetcarpet
Apr17-05, 01:10 PM
1)A spring with k = 40.0 N/m is at the base of a frictionless 30.0° inclined plane. A 0.50 kg object is pressed against the spring, compressing it 0.2 m from its equilibrium position. The object is then released. If the object is not attached to the spring, how far up the incline does it travel before coming to rest and then sliding back down? --I attached the picture, although it's not really neccessary due to the description above.--

I started by finding the Potential Energy of the spring:
U = 1/2kx^2
U = 1/2(40)(.2)^2 = .8 J
But, I have no idea where to proceed from that point forward. Can anyone help me out? I know that the answer is .33m, I simply can not understand how I am to arrive at that answer.

quasar987
Apr17-05, 01:31 PM
It's actually -0.8J. This amount will be transfered to kinetic energy after the block is released. It will then go up the ramp, where it will have to fight against the gravitational force, hence its kinetic energy will be transfered to gravitationnal potential energy. Knowing that U = mgh, you get the height it will attain.