Roller Coaster-Gravitational potential

In summary, the roller coaster car has a mass of 3,000kg and moves from point A to point C with a height difference of 35m and 23m respectively. The gravitational potential energy at these points is 1.029E+6J and 676,200J, respectively. The kinetic energy and speed at point B are 1.029E+6J and 26.19m/s, while at point C they are 352,973J and 15.34m/s. When taking into account friction, the work done by friction is 428,874J. This is found by using the work-energy theorem and calculating the difference in kinetic energy between point B and C.
  • #1
Dannystu
16
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?
 
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  • #2
Everything seems correct to me. At least, the equations are.. :wink:
Didn't do the calculations but I reckon you're right.
 
  • #3
Okay thanks! Could you possibly push me in the right direction on how to approach part d?
 
  • #4
d) Hint: What does the work-energy theorem say?
 
  • #5
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?
 
  • #6
anyone?
 

Related to Roller Coaster-Gravitational potential

What is the concept of gravitational potential in relation to roller coasters?

The concept of gravitational potential in relation to roller coasters refers to the potential energy that is stored in a roller coaster due to its height above the ground. This potential energy is converted into kinetic energy as the roller coaster moves downhill, providing the thrilling and fast-paced ride that we experience.

How is the gravitational potential energy of a roller coaster calculated?

The gravitational potential energy of a roller coaster can be calculated by multiplying the mass of the roller coaster by the acceleration due to gravity (9.8 m/s^2) and the height of the roller coaster above the ground. This calculation can also be used to determine the minimum height required for a roller coaster to complete a loop or other inversions.

What factors affect the gravitational potential energy of a roller coaster?

The gravitational potential energy of a roller coaster is affected by the mass of the roller coaster, the height of the roller coaster above the ground, and the acceleration due to gravity. In addition, the speed and shape of the roller coaster's track can also impact the amount of gravitational potential energy present.

How does the concept of gravitational potential impact the design and safety of roller coasters?

The concept of gravitational potential is a crucial factor in the design and safety of roller coasters. Designers must carefully consider the height, speed, and shape of the track to ensure the roller coaster has enough gravitational potential energy to complete its course without causing harm to riders. Additionally, safety measures such as seat belts and restraints are put in place to prevent riders from being thrown out of the roller coaster due to the force of gravity.

Why do roller coasters often have steep drops and sharp turns?

Roller coasters often have steep drops and sharp turns because they allow for a large change in gravitational potential energy. As the roller coaster travels down a steep drop, it gains a significant amount of kinetic energy, which can then be used to propel it through sharp turns and other thrilling elements. These design elements also add to the excitement and thrill of the ride for riders.

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