Answer:
The angle above horizontal is 13.3°.
Explanation:
The angle can be calculated with Newton's third law:
[tex] \Sigma F = ma [/tex]
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:
[tex] mgsin(\theta) = ma [/tex]
[tex] gsin(\theta) = a [/tex] (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:
[tex] v_{f}^{2} = v_{0}^{2} + 2ad [/tex]
Where:
[tex]v_{f}[/tex]: is the final speed = 3.00 m/s
[tex]v_{0}[/tex]: is the initial speed = 0 (the block starts from rest)
d: is the distance traveled = 2.00 m
The acceleration is:
[tex] a = \frac{v_{f}^{2}}{2d} = \frac{(3.00 m/s)^{2}}{2*2.00 m} = 2.25 m/s^{2} [/tex]
Finally, the angle is (equation 1):
[tex] \theta = sin^{-1}(\frac{a}{g}) = sin^{-1}(\frac{2.25 m/s^{2}}{9.81 m/s^{2}}) = 13.3 [/tex]
Therefore, the angle above horizontal is 13.3°.
I hope it helps you!
We have that for the Question "At what angle above horizontal is the inclined plane tilted"
It can be said that
From the question we are told
mass of the block m = 6 kg
distance traveled d = 2 m
speed of the block v = 3 m/s
Generally the equation for conservation of energy is mathematically given as
[tex]mgh = \frac{1}{2} mv^2[/tex]
so, height of the slope
[tex]h = \frac{v^2}{2g}\\\\= \frac{3^2}{2*9.8}\\\\= 0.459 m[/tex]
From Pythagorean theorem
[tex]sin\theta = \frac{0.459}{2}\\\\\theta = sin^{-1}\frac{0.459}{2}\\\\= 13.2^o[/tex]
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