A crate begins sliding down a frictionless incline of 25.0o. If the objects begins sliding from rest and takes 1.50 sec to slide down the incline, what is the speed of the crate as it leaves the incline? How long is the incline?
To find the speed of the crate as it leaves the incline, we first need to find the acceleration of the crate down the incline.
The acceleration of an object on an incline can be calculated using the following formula:
a = g * sin(θ)
Where:
a = acceleration
g = acceleration due to gravity (approximately 9.8 m/s^2)
θ = angle of the incline (in this case, 25.0 degrees)
Let's plug in the values:
a = 9.8 m/s^2 * sin(25.0 degrees)
Now we can calculate the acceleration:
a = 9.8 m/s^2 * 0.4226
a ≈ 4.143 m/s^2
Next, we can use the equation of motion for an object in linear motion to find the final velocity of the crate as it leaves the incline. The equation is:
v = u + at
Where:
v = final velocity (the speed of the crate as it leaves the incline)
u = initial velocity (in this case, the crate starts from rest, so u = 0)
a = acceleration (4.143 m/s^2, as we calculated earlier)
t = time (1.50 s, as mentioned in the question)
Plugging in the values:
v = 0 + 4.143 m/s^2 * 1.50 s
v ≈ 6.2145 m/s
Therefore, the speed of the crate as it leaves the incline is approximately 6.2145 m/s.
To find the length of the incline, we can use the equation of motion for an object in linear motion, where the displacement can be calculated using the following equation:
s = ut + (1/2)at^2
Since the initial velocity is 0 (the crate starts from rest), the equation simplifies to:
s = (1/2)at^2
Plugging in the values:
s = (1/2) * 4.143 m/s^2 * (1.50 s)^2
s ≈ 9.28575 m
Therefore, the length of the incline is approximately 9.28575 meters.