Dynamics of box moving in an inclined plane?

Question

The 50kg crate is projected down a 30 degree incline with respect to the horizontal with initial speed of 8m/s. Determine the time required for the crate to come to rest and the corresponding x traveled.

The coefficient of kinetic friction is 0.4

I kept getting incorrect answers.

The correct answer is t=6.39 s x=25.6m

Answer 1

Without answering your question explicitly, here are a few hints:

1. The acceleration of the crate is due to two forces: part of the weight, which is trying to speed it up; and the friction force, which is trying to slow it down.  It's the difference between these two forces that determines the acceleration rate (actually a deceleration, or negative acceleration, in this case).

2. The part of the weight that's pulling the crate down the ramp is related to the angle of the ramp (30 degrees).  Draw a picture of the crate on the ramp, then draw an arrow for the weight.  Then resolve the weight into a component along the ramp, and one perpendicular to (pushing down on) the ramp.  You'll use trigonometry to determine how much of the weight is pushing along the ramp.

3. The friction force is a function of the friction coefficient and the normal force between the crate and the ramp.  The normal force is the part of the weight that's perpendicular to the ramp.  It's not the whole weight; again, you'll use trig to figure out how much.

4. Once you have the friction force acting in one direction and the part of the weight pushing in the other direction, you can calculate the acceleration (actually deceleration) rate of the crate, which will tell you how long it takes to decelerate from 8m/s to zero.

5. Once you know the time for the crate to stop, you can calculate the distance from the position equation.

Hope this helps, without solving the problem for you.