A crate of oranges slides down an inclined plane without friction. If it is released from rest and reaches a speed of 4.883 m/s after sliding a distance of 2.01 m, what is the angle of inclination of the plane with respect to the horizontal? Please note: Give your answer in degrees!
Note
v = a t
the AVERAGE speed is 2.4415 m/s
so t = 2.01 meters/2.4415 m/s = .823 seconds
4.883 = a (.823)
a = 5.93 m/s^2
F = m a
m g sin A = m (5.93)
9.81 sin A = 5.93
sin A = 5.93/9.81
solve inverse sin for A
ý did 5,93/9,81 = 0,604485219.... but it is false... what is the answer exactly
ok ok ok ok ý did. thanks... :D:D
To find the angle of inclination of the plane, we can use the formula for the acceleration of an object sliding down an inclined plane without friction:
acceleration = g * sin(θ)
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and θ is the angle of inclination of the plane.
We can calculate the acceleration using the following formula:
acceleration = (final_velocity^2 - initial_velocity^2) / (2 * distance)
Substituting the given values into the equation:
acceleration = (4.883^2 - 0^2) / (2 * 2.01)
acceleration = 23.846 / 4.02
acceleration ≈ 5.933 m/s^2
Now we can rearrange the first equation to find the angle:
θ = arcsin(acceleration / g)
θ = arcsin(5.933 / 9.8)
To solve for θ, we can use a scientific calculator or an online trigonometric calculator to find the inverse sine (arcsin) of the value 5.933 / 9.8.
Evaluating this inverse sine, we find that θ ≈ 38.40 degrees.
Therefore, the angle of inclination of the plane with respect to the horizontal is approximately 38.40 degrees.