Forces (Just need it checked)

1. Dec 9, 2011

tarheels88

Question:
A crate of oranges slides down an inclined plane without friction. If it is released from rest and reaches a speed of 5.832 m/s after sliding a distance of 2.29m, what is the angle of inclination of the plane with respect to the horizontal?

I first set up a free body diagram then I discovered that finding the acceleration in the x direction would be the best idea.

I have the question down to a=7.42 m/s^2 which is correct for the acceleration, but I am getting stuck on where to go next. The formula I used is:

v^2= Vi^2+2a(x-xi) and that gave me the 7.42 m/s^2. The book says the angle is 41 degrees, but I don't know how to take the acceleration and determine the angle just from the acceleration. Anyone have some pointers?

2. Dec 9, 2011

ehild

Think: What forces act on the crate? What force accelerates the crate down the slope?

ehild

3. Dec 9, 2011

MrB8rPhysics

Thenacceleration along the slope is a component of g acting at an angle.... that should help. I reccomend drawing a nice big diagram of the situation with forces on it and looking for a nice right angled trianlge....

4. Dec 9, 2011

tarheels88

I did this but the problem never gave a mass. The angle is my desired result. Gravity is acting on the y direction F=mg. So I found acceleration in the x direction so...can I find the angle with the acceleration known even if the mass is not given?

5. Dec 9, 2011

MrB8rPhysics

You have 2 sides to a triangle - g of 9.81 in a vertical sense, and the component along the slope you calculated earlier. Could some trig be applied here?

6. Dec 10, 2011

ehild

Of course you can. You got a, the acceleration in the x direction. That corresponds to a force F=ma along the slope, red arrow in the picture, and that force is the parallel component of gravity (G, in blue). The arrows make a right triangle (yellow) with an angle α, equal to the inclination angle of the slope. You need only remember the definition of sine, cosine, tangent....

ehild

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7. Dec 10, 2011

lfcprg

I get the angle as 49.3 degrees to 3 s.f. Correct?

8. Dec 10, 2011

MrB8rPhysics

Possibly you have found the wrong the angle in the triangle there...

9. Dec 10, 2011

ehild

It has to be correct. The acceleration along the slope due to gravity is gsin(α). The acceleration is 7.426 ms-2, that corresponds to 49.2°, instead of 41.

ehild