Simple Inclined Plane Question.

[queenoftheclouds]

Sorry to bother with this stupid question, but I was hoping for some clarification.

Problem: A box starts from rest and slides down 40m down a frictionless inclined plane. The total vertical displacement of the box is 20m. How long does it take for the block to reach the end of the plane?

Answer: 4s

I understand how to get the answer if I take the x direction as mgsintheta (going down the plane), but my question is... why can't I solve the problem using the y displacement and the constant acceleration equation?

For example: y(final) - y(initial) = v(initial)t + 0.5at^2

-20 = (0.5)(-10)t^2

The answer I get using this method would be t=2s. What did I do wrong here?

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[doctor.octopus]

You're using the wrong value for a The box isn't accelerating at the rate of gravitational acceleration in the y-direction.

In the x-direction, the box is accelerating at a rate of 5m/s, or half of gravity, along the distance of 40m. We know this because:
Fgx = mgsintheta, and
F = ma, so in this case,
a = gsintheta. Thus, down the inclined plan, a is equal to g/2, or 5m/s.

We can intuit that this value for the y-direction should be even slower, since the time must be the same, and the distance is shorter. Since the y-distance is 20m, and this is half of 40m, this acceleration is equal to 2.5m/s.

The key here is that you need to relate movement along the vertical axis to the movement along the inclined plane, since it's the incline that's dictating the fact that acceleration is less than gravity. The calculations you posted weren't taking into account the inclined plane.

I think you're maybe thinking of energies, where we can use either the x or y direction, because energy is conserved

-DocOctopus