# Roller Coaster

**Topics:**Potential energy, Energy, Force, Introductory physics, Classical mechanics /

**Pages:**2 (288 words) /

**Published:**Jul 11th, 2013

E1=E2

Eg+Ek= Eg1+Ek1 mgh+1/2mv2= mgh1+1/2mv21

(1250)(9.8)(10)+1/2(1250)(v)2=(1250)(9.8)(65)+1/2(1250)(2.5)2

v= 32.9 m/s

Therefore you need 200003.9 watts of power to raise the cars to the top of the 1st hill with the speed of 32.9 m/s

Therefore you need 200003.9 watts of power to raise the cars to the top of the 1st hill with the speed of 32.9 m/s

Energy= 800156 J

P= E/T = 800156J/40s = 200003.9 W

b)

E1-Wf =E2

The expected speed of the cars is 8.5 m/s. Friction did 608000 J of work on the cars from the top of the hill to the braking zone.

The expected speed of the cars is 8.5 m/s. Friction did 608000 J of work on the cars from the top of the hill to the braking zone.

Eg+Ek-Wf= Eg1+Ek1 mgh+1/2mv2-Wf= mgh1+1/2mv21

Wf= mgh+1/2mv2- mgh1-1/2mv21 = (9.8)(65)+1/2(2.5)2-(9.8)(12)-1/2(8.5)2 = 608000 J

c)

Wf=Ff x d

The average Frictional Force was 844.4 N.

The average Frictional Force was 844.4 N.

Ff = Wf/d =608000J/720m = 844.4 N

Assumptions

* Acceleration of gravity is 9.8 m/s2 * Moving in direction of Force ( cos=0) * Sitting in middle cart. * Start 10 m off the ground “loading area” * Air resistance isn’t a huge factor

Physics

* The roller coaster is always moving (Kinetic Energy) * Gravitational Potential Energy is constant * Friction of Roller