Final answer:
The question pertains to physics, focusing on the acceleration of a block down a frictionless inclined plane and the forces acting on it. If the ramp makes an angle of 30° with the horizontal, the block's acceleration down the ramp will be 4.9 m/s², and the force of the ramp on the block will be approximately 16.97 N. To move the block up the ramp at a constant velocity, a force of 9.8 N needs to be applied upward along and parallel to the ramp.
Step-by-step explanation:
The question is derived from a physics concept regarding the dynamics of objects on an inclined plane. Specifically, it addresses finding the acceleration of a block down a ramp and the forces involved when a block is placed on a frictionless inclined plane.
Example Answer:
For a block on a frictionless inclined plane, the acceleration of the block, a, can be found using the equation a = g × sin(θ), where g is the acceleration due to gravity (9.8 m/s²) and θ is the angle of the incline. For a 30° incline,
a = 9.8 m/s² × sin(30°) = 9.8 m/s² × 0.5 = 4.9 m/s²
This is the block's acceleration down the ramp. The force of the ramp on the block, also known as the normal force, N, can also be calculated using N = m × g × cos(θ), where m is the mass of the block. With a mass of 2.0 kg,
N = 2.0 kg × 9.8 m/s² × cos(30°) ≈ 16.97 N.
To move the block with a constant velocity up the ramp, an upward force equal to the component of the weight of the block along the ramp should be applied, which would be F = m × g × sin(θ). Thus, this force must be
F = 2.0 kg × 9.8 m/s² × sin(30°) = 2.0 kg × 9.8 m/s² × 0.5 = 9.8 N.