If a roller-coaster car has 40,000 J of gravitational potential energy when at rest on the top of a hill, how much kinetic energy does it have when it is ½ of the way down the hill?

Work Shown:

PE = potential energy

KE = kinetic energy

m = mass

g = acceleration of gravity

h = height of the object

PE = m*g*h = 40,000 joules

If the car is halfway down the hill, then that means the height of the car is h/2 instead of h

We can then say:

m*g*(h/2) = (m*g*h)/2 = (40,000)/2 = 20,000 J

is the amount of potential energy at this location. The amount of kinetic energy must be the remaining amount that adds to this, to get 40,000 J again.

Put another way: The 40,000 J of PE to start off with, when KE = 0, means we have a total energy of 40,000 J

So,

PE+KE = total energy

PE+KE = 40,000 J

20,000 J + KE = 40,000 J

KE = (40,000 J) - (20,000 J)

KE = 20,000 J when the car is halfway down the hill.

the tires are rotating at 0.434 rpmlinear speed=45 m/sdiameter=0.550 mradius=0.275 mnow v= rωω= v/rω=45/0.275ω=163.6 rad/sω=163.6 rad/s (1rev/2π rad) (60s/1m)ω=1563 rpm