Final velocity of a ball down a ramp.

[HHH]

Homework Statement

If a ball is dropped down a ramp of any shape, will the acceleration always be 9.8 assuming there is no friction.

Homework Equations

I know you can use mgh = 1/2mv^2 to solve for velocity, but can you use vf^2 = vi^2 + 2(a)(d)

The Attempt at a Solution

If you use both equations you get the same answer for final velocity at the bottom of the ramp. However is the acceleration always 9.8 no matter what the shape of the ramp is?

 

[haruspex]

I don't see any attempt to answer the question here. Think about forces.

 

[HHH]

Does it have to do with the normal force decreasing the force of gravity?

 

[haruspex]

Does it have to do with the normal force decreasing the force of gravity?

Well, it doesn't decrease it, but it does partially oppose it.

 

[HHH]

so the acceleration would not be 9.8? if not why can you use both equations?

mgh = 1/2mv^2
(2)x(9.8)x(3) = 1/2 (2) v^2
v = 7.6m/s

vf^2 = vi^2 + 2ad
vf^2 = 0 + 2(9.8)(3)
v = 7.6m/s

 

[haruspex]

so the acceleration would not be 9.8? if not why can you use both equations?

mgh = 1/2mv^2
(2)x(9.8)x(3) = 1/2 (2) v^2
v = 7.6m/s

vf^2 = vi^2 + 2ad
vf^2 = 0 + 2(9.8)(3)
v = 7.6m/s

Ignoring the fact that it's a ball for the moment, just treating it as a zero friction ramp, those (that) equation only tells you the final velocity. It doesn't tell you how long it takes to get there. The final velocity is the same whether dropped straight down or going down a ramp. It only depends on the PE lost. But it takes longer down the ramp, so the acceleration is less.

If it is a rolling ball, that starts off not rotating, then there is another issue. At the end some of the KE will be in rotational energy, so the linear velocity will be less.

 

[HHH]

Thank you.