Kinetic and Potential Energy on a roller coaster

  • #1

Homework Statement


A roller coaster starts at the top of hill that is 10 m high. If it is to barely make it to the top of a second hill that is 20 m high, how fast must the initial speed of the roller coaster be? Assume that the roller coaster is frictionless.


Homework Equations


KE = 0.5mv^2
PE = mgh

The Attempt at a Solution


mgh=0.5mv^2
gh = 0.5v^2
v = sqrt(2gh)
v = sqrt(2*9.81*20)
v = 20 m/s needed at bottom of second hill.

mgh=0.5mv^2
gh = 0.5v^2
v = sqrt(2gh)
v = sqrt(2*9.81*10)
v = 14 m/s at bottom of hill if initial velocity at top of first hill is 0m/s.

20 m/s - 14 m/s = 6 m/s

6 m/s is the velocity of the roller coaster at the top of the first hill.

Does that look right?
 

Answers and Replies

  • #2
6,054
391
No, this is not right. Check it this way: compute KE with v = 6 m/s and PE with h = 10 m; the sum is the total energy initially. It must be equal to the total energy finally, which is KE with v = 0 m/s and PE with h = 20 m.
 
  • #3
Hm, okay. I tried setting the total energy at the top of hill one equal to the total energy at the top of hill two.
0.5m(v1)^2+mg(h1) = 0.5m(v1)^2+mg(h2)
0.5(v1)^2 + g(h1) = 0.5(v1)^2 + g(h2)
v = sqrt(2*g*(h2-h1))
v = sqrt (2*9.81*(20-10))
v = 14 m/s

Would that work?
 
  • #4
6,054
391
The result is correct, but the first two equations are not correct, having the same v1 symbol on both sides.
 
  • #5
Oh, yeah... I meant (v2) on the right side of the equations. That was simply a typo.
 
  • #6
6,054
391
Very well then.
 

Related Threads on Kinetic and Potential Energy on a roller coaster

  • Last Post
Replies
10
Views
6K
  • Last Post
Replies
18
Views
30K
Replies
3
Views
2K
Replies
12
Views
7K
Replies
1
Views
2K
Replies
2
Views
2K
  • Last Post
Replies
4
Views
981
Replies
1
Views
1K
  • Last Post
Replies
19
Views
4K
Replies
2
Views
911
Top