An 800kg roller coaster starts from top of a 45m hill with a velocity of 4m/s. the car travels to the bottom, through a loop, and continues up the next hill. the end of the roller coaster has a level surface that is 4m off the ground. assume there is no friction on the roller coaster ridea and energy is concerved.
q. how tall is the hill if the roller coaster is traveling at 10m/s at the top of it?
q. what will be the velocity of the roller coaster at the top of the loop?

Let's start with the total amount of energy available for the whole scenario:
Some kind of machine gave the coaster a bunch of potential energy by
dragging it up to the top of a 45m hill,and that's the energy is has to work with.

Potential energy = (M) (G) (H)  =  (800) (9.8) (45) = 352,800 joules

It was then given an extra kick ... enough to give it some kinetic energy, and
start it rolling at 4 m/s.

Kinetic energy = (1/2) (M) (V)² = (1/2) (800) (4)² = 6,400 joules

So the coaster starts out with (352,000 + 6,400) = 359,200 joules of energy.

There's no friction, so it'll have that same energy at every point of the story.


Skip the loop for a moment, because the first question concerns the hill after
the loop.  We'll come back to it.

The coaster is traveling 10 m/sat the top of the next hill. Its kinetic energy is

(1/2) (M) (V)² = (400) (10)² = 40,000 joules.

Its potential energy at the top of the hill is (359,200 - 40,000) = 319,200.

PE = (M) (G) (H)

319,200 = (800) (9.8) (H)

H = (319,200) / (800 x 9.8) = 40.71 meters


Now back to the loop:

You said that the loop is 22m high at the top. The PE up there is

PE = (M) (G) (H) = (800) (9.8) (22) = 172,480 joules

So the rest is now kinetic. KE = (359,200 - 172,480) = 186,720 joules.

KE = (1/2) (M) (V)² = 186,720

(400) (V)² = 186,720

V² = 186,720 / 400 = 466.8

V = √466.8 = 21.61 m/s


Now it looks like there should be another question ... that's why they
bothered to tell you that the end is 4m off the ground. They must
want you to find the coaster's speed when it gets to the end.

At 4m off the ground,  PE = (M) (G) (H) = (800) (9.8) (4) = 31,360 joules.

The rest will be kinetic.  KE =  (359,200 - 31,360) = 327,840 joules

KE = (1/2) (M) (V)² = 327,840

400 V² = 327,840

V² = 327,840 / 400 = 819.6

V = √819.6 = 28.63 m/s at the end


If the official answers in class are a little bit different from these,
it'll be because they used some different number for Gravity.
I used '9.8' for gravity, but very often, they use '10' .

If the official answers in class are way way different from these,
then I made one or more big mistakes somewhere.  Sorry.

The car would move to the right with 150 n of force. the 150 n of force from friction and air resistance slows the car from moving to the right with 300 n of force.