A block starts from rest at the top of a 2.00-m long, 16.3 -degree frictionless

incline.
(a) What is the acceleration (m/s2 ) of the block?
(b) How fast (m/s) is it going at the bottom?

(a) It equals the component of g along the direction of motion.

That would be 9.8 sin16.3 m/s^2

(b) The vertical drop is
H = 2.00 sin 16.3 = 0.561 m

Conservation of energy tells you that, with no friction,
V^2 = 2 g H after falling a vertical distance H

To solve this problem, we can use the principles of energy conservation and the equations of motion.

(a) Acceleration of the block:
- We know the distance traveled (2.00 m) and the angle of the incline (16.3 degrees). To find the acceleration, we need to resolve the gravitational force perpendicular and parallel to the incline.
- The gravitational force can be decomposed into two components: F_parallel = m * g * sin(theta) and F_perpendicular = m * g * cos(theta), where m is the mass of the block, g is the acceleration due to gravity (approximately 9.8 m/s^2), and theta is the angle of the incline.
- The net force parallel to the incline is responsible for the acceleration of the block.
- F_net_parallel = m * a_parallel, where a_parallel is the acceleration of the block parallel to the incline.
- Since the incline is frictionless, there is no force opposing the motion parallel to the incline, so F_net_parallel = m * a_parallel = m * g * sin(theta).
- Rearranging the equation, we get a_parallel = g * sin(theta).
- Now, we can substitute the values: a_parallel = 9.8 m/s^2 * sin(16.3 degrees).
- Calculating this will give us the acceleration of the block.

(b) Speed of the block at the bottom:
- To find the speed at the bottom of the incline, we can use the equations of motion.
- We know the initial velocity is zero since the block starts from rest. However, we need to find the final velocity at the bottom of the incline.
- Using the equation v^2 = u^2 + 2as, where v is the final velocity, u is the initial velocity, a is the acceleration, and s is the distance traveled, we can find the final velocity.
- Rearranging the equation, we get v^2 = 2as, and solving for v gives v = sqrt(2as).
- The distance traveled down the incline is 2.00 m, and the acceleration parallel to the incline is the one we calculated in part (a).
- Substituting these values, we can find the velocity at the bottom of the incline.

Hope this helps! If you need any further assistance, feel free to ask.

To find the acceleration of the block on the incline, we need to break down the force acting on the block. There are two forces at play: the gravitational force (mg) and the force component parallel to the incline (mg sinθ), where θ is the angle of the incline.

(a) To find the acceleration, we can use Newton's second law: F = ma. The net force on the block is the component of the gravitational force parallel to the incline, i.e., Fnet = mg sinθ. Therefore, we can write:

Fnet = ma

mg sinθ = ma

Simplifying the equation, we find:

a = g sinθ

where g is the acceleration due to gravity (approximately 9.8 m/s²) and θ is the angle of the incline (16.3 degrees in this case). Plugging in the values, we get:

a = (9.8 m/s²) * sin(16.3°)

Calculating this, we find the acceleration to be:

a ≈ 2.6 m/s²

(b) Now, to find the velocity of the block at the bottom of the incline, we can use the equation of motion:

v² = u² + 2as

where v is the final velocity, u is the initial velocity (in this case, 0 m/s as the block starts from rest), a is the acceleration we just calculated, and s is the distance traveled (2.00 m in this case, as it is the length of the incline).

Plugging in the values, we get:

v² = (0 m/s)² + 2 * (2.6 m/s²) * (2.00 m)

Simplifying the equation, we find:

v² = 10.4 m²/s²

Taking the square root of both sides, we get:

v ≈ 3.2 m/s

Therefore, the velocity of the block at the bottom of the incline is approximately 3.2 m/s.