# Inclined Plane and sliding block

## Homework Statement

The velocity of a 2.10 kg block sliding down a frictionless inclined plane is found to be 1.13 m/s. 1.00 s later, it has a velocity of 4.13 m/s. What is the angle of the plane with respect to the horizontal?

2. The attempt at a solution
I drew a diagram and everything...but I have never had to solve for an angle so I am not sure what formula to even start with.

Okay I have..
Vo = 1.13 m/s
Vf = 4.1 m/s
t = 1.00s
w = 20.6 N
m = 2.1 kg

I drew a diagram having the block sliding down the incline toward the origin and on the x-axis and I split the Normal into x and y components, can I do that? The only other force I have put on the block is the weight force. Is that right?

mass/weight has nothing to do in this problem.

you know what the acceleration is (g). But since it's at some angle (theta) it's not 100% g.

so a=sin or cos (theta)?

and then can you plug that into an equation that involves v0, vf, and time and solve for sin or cos (theta) and then you do arcsin or arccos to get the theta.

so acceleration = cos theta

no sin theta because it's going to be some y value, instead of being 90* straight up.
it's inclines somewhat so it's sin theta.

so I got a = 3 using vf = vo + at

yes but now you need to replace a by what a is when it's not a straight 90* or 0*.

so when you have that you need to change to a=9.8sin theta.

and then you solve for theta.

Also did you found to get a=3? don't round it off till you solve finally for theta or your answer could be off by too much.

I got 3 from (4.13m/s-1.13m/s) x 1.00s= 3 exactly.
so 3.00 = 9.8 sin theta
3.00/9.8 = sin theta
theta = 17.81?

Thank you very much!

yes theta = 17.81 degrees.

and sry saw 4.1 m/s on your post #2 not #1.

oh sorry!!