Roller Coaster-Gravitational potential

  • Thread starter Thread starter Dannystu
  • Start date Start date
  • Tags Tags
    Potential Roller
AI Thread Summary
The discussion focuses on a roller coaster physics problem involving gravitational potential energy, kinetic energy, and the effects of friction. The gravitational potential energy at points A, B, and C is calculated, revealing values of 1,029,000 J at A, 0 J at B, and 676,200 J at C. The kinetic energy at point B is determined to be 1,029,000 J, with a calculated speed of 26.19 m/s, while the speed at point C is found to be 15.34 m/s with a kinetic energy of 352,973 J. The discrepancy between the expected speed at point B and the measured speed indicates the presence of friction, leading to a calculation of the work done by friction as 428,874 J. The work-energy theorem is applied to analyze the energy changes throughout the roller coaster ride.
Dannystu
Messages
16
Reaction score
0

Homework Statement



The figure shows a roller coaster car mass m= 3,000kg, on a portion of a roller coaster ride. The height difference between points A and B is 35m. The height difference between B and C is 23m. The car starts from rest at point A. Take y=0 point to be point B. For parts a.,b., and c., assume that the roller coaster track is frictionless.



Homework Equations


a.) Compute the gravitational potential energy of the car at points A, B, and C.
b.) Compute the kinetic energy of the car at point B. Compute its speed there. What physical principle did you use to do this calculation?
c.) compute the kinetic energy and speed of the car at point C.
d.) Parts a., b., and c. assume that the track is frictionless. However, the measured speed of the car at point B is found to be V=20m/s. This is less than the speed that you should have computed in part b. This means that friction cannot be neglected. In this case, how much work is done by friction when the car moves from point A to point B? What physical principle did you use to compute this work?



The Attempt at a Solution



a.) gravitational potential energy =m*g*h

Point A= 3000*9.8*35= 1.029E+6J

Point B= 3000*9.8*0= 0J

Point C= 3000*9.8*23= 676,200J

b.) E=(1/2)m*v^2+m*g*h+(1/2)kx^2

At point A, there is no kinetic energy and no elastic energy
At point B, there is no gravitational potential energy or elastic energy
And since Ea=Eb
m*g*h=(1/2)m*v^2

Vb=26.19 m/s

KE=(1/2)m*v^2
KE= 1.029E+6J, same as the gravitational potential energy at point A.

C.) There is gravitational potential energy and kinetic energy at point C, so:

(1/2)m*v^2=(1/2)m*v^2+m*g*h

Vc= 15.34 m/s

KE= 352,973 J

d.) I wanted to make sure I was doing everything right before I go on lol. Can anyone verify my answers thus far?
 
Physics news on Phys.org
Everything seems correct to me. At least, the equations are.. :wink:
Didn't do the calculations but I reckon you're right.
 
Okay thanks! Could you possibly push me in the right direction on how to approach part d?
 
d) Hint: What does the work-energy theorem say?
 
Work-energy theorem:

W=f*d*cos(theta)

and

W= KE final- KE initial

so:

(1/2)*3000*(26.19)^2- (1/2)* 3000*(20)^2 = 428,874J

Is that right?
 
anyone?
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top