Physics Springs Question?

Question

A spring with k = 10.0 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 0.1 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?
____m

Answer 1

First, find the total energy in the system, and remember that energy is always conserved.
At the beginning there is potential energy in the spring.
PE=(1/2)kx^2. So PE=(1/2)*10N/m*(0.1)m^2.
PE= 0.05J
All that energy is converted to the kinetic energy of the object. This calculation is unnecessary, as that kinetic energy is converted to gravitational potential energy of the block.
GPE=mgh. So 0.05J=(0.5kg)*9.8*h. h=0.01m

So the block will be .01m high which means that it slides
0.01/sin(30)= 0.02m

answer 0.02m

Answer 2

This is solved using conservation of energy

The energy stored in the spring will be released and result in KE which gets converted to PE. The only part that is a bit unclear is where do you want to measure from: the compressed point or the equilibrium point?
The solution from the compressed point is:
.5*10*0.1^2=d*sin(30)*0.5*9.8

solve for d, which is the distance up the incline

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