How Far Will the Block Travel Up the Incline After Spring Release?

[Syrus]

Homework Statement

Consider the following situation:

A 2.0 kg block is placed against a compressed spring on a frictionless 30° incline. The spring
of force constant 19.6 N/cm is compressed 20 cm and then released. How far up the incline will the block go before coming to rest?

Homework Equations

The Attempt at a Solution

The suggested (but perhaps naive) solution to this problem is to calculate the potential energy stored in the spring when it is compressed and set this value equal to the change in gravitational potential energy (see, for instance, http://www.physics.udel.edu/~jim/PHYS207_10J/Homework%20Solutions/Hwk7sol.pdf ). However, I see two fundamental problems with this approach:

First, suppose our system is composed of the block, spring, ramp, and Earth. Then at its highest point (as with its initial condition when it compresses the spring), the block has no kinetic energy. Let the release point be the reference point for gravitational potential energy (potential energy = 0) and the unstretched equilibrium length of the spring to be its potential energy reference point. What is to say that when the block reaches its highest point, the spring will be at its equilibrium position? If this is not the case, then it certainly can't be true that the entire spring potential energy may be converted to gravitational potential energy. What I see the problem to be is attributing the potential energy stored in the spring to the energy transferred to the ball. Is this not a problem, in general?

This problem is from an elementary text and, as such, may not be written paying attention to such details. Nonetheless, I have devised a solution which is independent of the abovementioned approach, but still poses some questions:

We may calculate the work done by the spring on the box as it is released (since we are given the force constant and initial compression of the spring) and set this value equal to the change in gravitational potential energy (in this case we take only the box, ramp, and Earth as our system). This equation holds since here an external force acts on our system. This approach also yields the *suggested solution. However, it doesn't seem obviously true that the spring stops acting (contacting) on the box as it passes through its equilibrium position (which is an underlying assumption of this method)?

 

[Syrus]

Any advice?

 

[CWatters]

I think your initial approach is right. Apply conservation of energy...The block continues up until the PE stored in the spring has all been converted to potential energy in the block.

I think the problem is that simple but you are right that in the real world it might not be quite like that..

What is to say that when the block reaches its highest point, the spring will be at its equilibrium position

Why does that matter? The block is not fixed to the spring just placed against it. Once the spring reaches it's rest position it's delivered all it's PE to the block.

In the real world the spring has mass and the end of the spring will have velocity and KE as it reaches and passes the rest point. In other words not all the PE ends up in the block. The question doesn't state that the spring is mass less but the mass isn't given so you can worry about that sort of thing. A massless spring can do things like stop instantly :-)

 

[Syrus]

Ah, so the fundamental, simplifying assumption is that the spring instantly come to rest upon reaching its equilbrium length. Perhaps this should be stated in the problem, no?