Question

If a 20kg ball is sitting at rest on the top of a 100m hill, what it the velocity of the ball when it rolls down the hill and is 10m from the bottom? Show, at least, your final calculation.

Answer 1

Step-by-step explanation:

This is the First Law of Thermodynamics that says that energy cannot be created nor destroyed. Which means, to us, that the ball has a specific amount of energy available to it, regardless of where it is in its travels, and this amount of energy will never be increased or decreased, that it will ust change its form. The Total Energy equation for this situation is

TE = PE + KE whih says that the total energy available to the system is equal to its Potential Energy plus its Kinetic Energy, and that Total Energy will never change. First we need to find the TE available. Looking at the first statement, we come into this problem and the ball is sitting still at the top of a hill. If the ball is sitting still, it has no Kinetic Energy, so all the energy is Potential. Therefore,

TE = PE + 0 and what we find here for the TE is the TE available throughout the trip. PE = mgh, so

TE = (20.0)(9.8)(100.0) I added in some sig dig but will still round to 2 because 1 does nothing for us.

TE = 2.0 × 10⁴ J. Now we move on to the next part of the problem, where the ball starts moving. The instant the ball starts moving, the PE starts to convert to KE, but as long as the ball still has the potential to fall to a lesser height, it also still has PE...long story short, it has both PE and KE at this point. And since TE = PE + KE, then

2.0 × 10⁴ = (20.0)(9.8)(10.0) +
`(1)/(2)(20.0)v^2` and we solve for v, simplifying some at first.

2.0 × 10⁴ = 2.0 × 10³ + 10.0v² and

`v=\sqrt{(2.0*10^4-2.0*10^3)/(10.0) }` so, to 2 sig fig,

v = 42 m/s

Get to know how to do this; it will save your sanity!!