A block lies on a smooth plane tilted at an angle è = 28.5° to the horizontal. Ignore friction.
(a) Determine the acceleration of the block as it slides down the plane.
1 m/s2
(b) If the block starts from rest 9.40 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?
2 m/s
To determine the acceleration of the block as it slides down the plane, we can use the following formula:
acceleration = g * sin(θ)
where g is the acceleration due to gravity and θ is the angle of the plane. In this case, θ = 28.5°.
(a) Determine the acceleration of the block as it slides down the plane:
acceleration = g * sin(θ)
acceleration = 9.8 m/s^2 * sin(28.5°)
acceleration ≈ 4.90 m/s^2
Therefore, the acceleration of the block as it slides down the plane is approximately 4.90 m/s^2.
To determine the speed of the block when it reaches the bottom of the incline, we can use the following formula of motion:
v^2 = u^2 + 2 * a * s
where v is the final velocity (speed), u is the initial velocity (which is 0 in this case since the block starts from rest), a is the acceleration down the incline, and s is the distance traveled down the incline.
(b) If the block starts from rest 9.40 m up the plane from its base, what will be the block's speed when it reaches the bottom of the incline?
We need to find v. Given:
u = 0 m/s (starting from rest)
a = 4.90 m/s^2 (from part (a))
s = 9.40 m (distance traveled down the incline)
Using the formula v^2 = u^2 + 2 * a * s, we have:
v^2 = 0^2 + 2 * 4.90 m/s^2 * 9.40 m
v^2 = 0 + 92.12 m^2/s^2
v ≈ √92.12 m^2/s^2
v ≈ 9.60 m/s
Therefore, the block's speed when it reaches the bottom of the incline will be approximately 9.60 m/s.