Solving for Angle of Inclined Plane: 2.3kg Block Sliding Down Frictionless Plane

[jemck]

Here is the problem: The velocity of a 2.3 kg block sliding down a frictionless inclined plane is found to be 1.12 m/s. 1.20 sec. later it has a velocity of 5.08 m/s. What is the angle of the plane with respect to the horizontal?

There is quite a bit of information here. I have gathered some of the vital details and will demonstrate what I have thus far, which may be correct or incorrect.

This is a Newton's Second Law Problem (I believe).
F=m*a
accel=v2-v1/t2-t1 = (5.08 - 1.12)/(1.2-0)= 3.3m/s^2
So, at this point I have mass and acceleration to work with.
I know that there are three forces acting on the block, so I have constructed a free body diagram.

Also, solving for Fgrav would require w=m*g. w=(2.3)*(3.3) therefore w=22.545 [N].

Most problems involving an inclined plane have theta given. However, I have not been able to find any problems similar to this one with theta unknown.

Any help or website resources would be greatly appreciated.

 

[Jameson]

You have a right triangle. You can express two of the sides in terms of acceleration (m/s^2). Use a trig inverse property to solve for the angle. ($$arcsin(x)$$)

 

[futb0l]

$$F = mg sin \theta = m\frac{v_f - v_i}{t}$$

That's enough info! :P