What is the acceleration of a block sliding down a frictionless ramp?

The force of gravity near the surface of the Earth is:

Fg = mg

g is the acceleration of a falling body near the surface of the Earth.

The above figure shows a block sliding down a frictionless ramp.

The component of the force of gravity in the direction of the block’s motion is:

c Fg = Fg sinθ = mg sinθ

The net force acting on the block is the component of the force of gravity in the direction of the block’s motion. We have:

F = c Fg

ma = mg sinθ

a = g sinθ

For a body that starts from rest and travels with uniform acceleration, we have the following equation:

v ² = 2ax

Falling bodies travel with uniform acceleration. For a falling body, we have the following equation:

v ² = 2gy

g is the acceleration of a falling body, y is how far the body has fallen. The displacement of the falling body is measured downward from the point where the body was dropped, so displacement is positive and acceleration is positive.

The block in the above figure has slid the distance z down a frictionless ramp. The block slides down the ramp with uniform acceleration.

The change in gravitational potential energy of the block is the same whether it falls the distance y or slides the distance z.

The kinetic energy of the block goes up the same amount as the gravitational potential energy of the block goes down. So the kinetic energy of the block is the same whether it falls the distance y or slides the distance z.

So the velocity of the block is the same whether it falls the distance y or slides the distance z.

The velocity of the block if it falls the distance y is:

v ² = 2gy

The velocity of the block if it slides the distance z is:

v ² = 2az

Setting these two equations equal to each other,

2az = 2gy

a = gy ⁄ z = g sin θ