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!

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To find the angle of inclination of the plane, we can use the equation for acceleration on an inclined plane:

a = g * sin(θ)

where
- a is the acceleration of the crate,
- g is the acceleration due to gravity (approximately 9.8 m/s²), and
- θ is the angle of inclination of the plane.

From the problem statement, we know that the crate slides down the inclined plane without friction. This means that the force of friction is zero and the net force acting on the crate is equal to its weight.

The net force can be calculated using Newton's second law:

F_net = m * a

where
- F_net is the net force acting on the crate,
- m is the mass of the crate.

Assuming the mass of the crate is not provided in the question, we can cancel out the mass by dividing both sides of the equation by m:

F_net / m = a

Since the net force is equal to the weight of the crate, we can write:

m * g * sin(θ) = m * a

The mass cancels out, leaving us with:

g * sin(θ) = a

Now we can substitute the given values:

g * sin(θ) = 4.883 m/s²

Rearranging the equation to isolate sin(θ):

sin(θ) = 4.883 m/s² / g

sin(θ) = 4.883 m/s² / 9.8 m/s²

sin(θ) ≈ 0.4988

Next, we need to find the angle whose sine is approximately 0.4988. This can be done by taking the inverse sine (also known as arcsine) of 0.4988:

θ ≈ arcsin(0.4988)

Using a calculator, we find that:

θ ≈ 29.4°

Therefore, the angle of inclination of the plane with respect to the horizontal is approximately 29.4°.