- #1
tarheels88
- 8
- 0
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?
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?