Answer:
The angle above horizontal is 13.3°.
Explanation:
The angle can be calculated with Newton's third law:
![\Sigma F = ma](https://tex.z-dn.net/?f=%20%5CSigma%20F%20%3D%20ma%20)
Where:
ΣF: is the forces acting on the object
m: is the mass of the object = 6.00 kg
a: is the acceleration of the object
The only force acting on the object is the weight since there is no friction, so:
(1)
Where:
θ: is the angle
g: is the acceleration due to gravity = 9.81 m/s²
We can find the acceleration from the following kinematic equation:
![v_{f}^{2} = v_{0}^{2} + 2ad](https://tex.z-dn.net/?f=%20v_%7Bf%7D%5E%7B2%7D%20%3D%20v_%7B0%7D%5E%7B2%7D%20%2B%202ad%20)
Where:
: is the final speed = 3.00 m/s
: is the initial speed = 0 (the block starts from rest)
d: is the distance traveled = 2.00 m
The acceleration is:
![a = \frac{v_{f}^{2}}{2d} = \frac{(3.00 m/s)^{2}}{2*2.00 m} = 2.25 m/s^{2}](https://tex.z-dn.net/?f=%20a%20%3D%20%5Cfrac%7Bv_%7Bf%7D%5E%7B2%7D%7D%7B2d%7D%20%3D%20%5Cfrac%7B%283.00%20m%2Fs%29%5E%7B2%7D%7D%7B2%2A2.00%20m%7D%20%3D%202.25%20m%2Fs%5E%7B2%7D%20)
Finally, the angle is (equation 1):
![\theta = sin^{-1}(\frac{a}{g}) = sin^{-1}(\frac{2.25 m/s^{2}}{9.81 m/s^{2}}) = 13.3](https://tex.z-dn.net/?f=%20%5Ctheta%20%3D%20sin%5E%7B-1%7D%28%5Cfrac%7Ba%7D%7Bg%7D%29%20%3D%20sin%5E%7B-1%7D%28%5Cfrac%7B2.25%20m%2Fs%5E%7B2%7D%7D%7B9.81%20m%2Fs%5E%7B2%7D%7D%29%20%3D%2013.3%20)
Therefore, the angle above horizontal is 13.3°.
I hope it helps you!