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Roller coaster car energy problem

  • Thread starter nns91
  • Start date
  • #1
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Homework Statement



1. A roller coaster car of mass 1500kg starts a distance H=23m above the bottom of a loop 15m in diameter. If friction is negligible, the downward force the rails on the car when it is upside down at the top of the loop is:
a. 4.6*10^4 N
b.3.1*10^4 N
c.1.7*10^4 N
d. 980N
e.1.6*10^3

2.A block of mass m is pushed up against a spring, compressing it a distance x and the block is then released. The spring projects the block along a frictionless horizontal surface, giving the block a speed v. The same spring projects a second block of mass 4m, giving it a speed 3v. What distance was the spring compressed in the second case ?

Homework Equations



E=(1/2)*k*x^2 + (1/2)*m*v^2 + mgh

The Attempt at a Solution



1. So I wrote out energy equation:

Initial: E=mgh=1500*g*23

Final: E= (1/2)*m*v^2 + mgh = (1500*v^2/2)+ 1500*15*g

then I set them equal and get 23*g= (1/2)*v^2 + 15g then I solve for v^2= 16*g

F=mv^2/2 = 16*9.81*1500 / 7.5 = 31392 N then I chose b. However, I am wrong. can you guys help me identify where did I do wrong ?


2.I also did the same

Initial: E= k*x^2 /2

Final: E= m*v^2 /2

I set them equal and get k*x^2= m*v^2

I substitute 4m for mass and 3v for speed and get k*x^2 = 4m*9v^2

I solved for x= squroot (4*m*9*v^2/k)

My answer was wrong too. I think because it contains k and they don't give k in the givens. How can I fix it ??
 

Answers and Replies

  • #2
590
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1. Only the force is asked, not energy.

2. Take compression of the spring as x1 and x2 for the two blocks and write down corresponding equations. Now relate x1 and x2.
 
  • #3
301
0


1. Right. I figured out the velocity by using energy and I used F=mv^2/r Am I wrong ?

2. I don't quite get what you mean
 
  • #4
301
0


I still got the wrong answer.
 

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