A block is projected up a frictionless inclined plane with initial speed v0 = 1.92 m/s. The angle of incline is è = 12.2°. (a) How far up the plane does it go? (b) How long does it take to get there? (c) What is its speed when it gets back to the bottom?
To find the answers to the given questions, we need to analyze the motion of the block on the inclined plane. We can break down the motion into two components: one along the incline and one perpendicular to it.
(a) To determine how far up the plane the block goes, we need to find the vertical displacement (height). We can use trigonometry to relate the angle of incline to the height.
The vertical displacement (height) can be found using the formula:
height = initial velocity * time * sin(angle of incline)
Given:
initial velocity (v0) = 1.92 m/s,
angle of incline (è) = 12.2°.
First, convert the angle from degrees to radians:
angle in radians = angle in degrees * (π/180)
angle in radians = 12.2° * (π/180) ≈ 0.213 rad
Next, we can substitute the values into the formula to find the height:
height = 1.92 m/s * time * sin(0.213 rad)
Now, since the block starts from rest at the top, the vertical displacement at the top must be zero. Therefore, the height is equal to zero.
0 = 1.92 m/s * time * sin(0.213 rad)
To solve for time, rearrange the equation:
time = 0 / (1.92 m/s * sin(0.213 rad))
time = 0
Therefore, the block does not travel any height up the plane.
(b) The time taken to reach the top of the plane is zero seconds.
(c) To find the speed when the block gets back to the bottom, we need to analyze the motion along the incline.
The speed when it gets back to the bottom is equal to the initial speed, as there is no friction on the inclined plane. Therefore, the speed is 1.92 m/s.