Law of conservation of Energy question.

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SUMMARY

The discussion centers on applying the law of conservation of energy to solve a roller coaster physics problem involving a 1000 kg coaster moving through various heights. The maximum speed at the lowest point (Point D) is calculated to be 42.4 m/s using the formula mgh = 0.5mv^2, where the height difference is 9m. The speed at Point E is determined to be 35.3 m/s, and the braking force required to stop the coaster within 5m is derived using F = ma and kinematic equations. Key insights emphasize the relationship between potential and kinetic energy in roller coaster dynamics.

PREREQUISITES
  • Understanding of the law of conservation of energy
  • Familiarity with gravitational potential energy (mgh)
  • Knowledge of kinetic energy (0.5mv^2)
  • Basic principles of kinematics (v^2 = u^2 + 2as)
NEXT STEPS
  • Study the derivation of the conservation of energy equation in mechanical systems
  • Learn about the effects of friction on motion and energy loss
  • Explore advanced kinematic equations for uniformly accelerated motion
  • Investigate real-world applications of energy conservation in amusement park rides
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of energy conservation in mechanical systems, particularly in the context of roller coaster dynamics.

ultimatejester
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I have a exam tomorrow so i will greatly appreciate anyone who helps me on this one. The question has to be solved using law of conservation of energy.

Q: A roller coaster starts from rest at point A and weighs 1000kg with the ppl in it and goes through the designed track. The height of the track changes 5 times. There is a pic but i don't know how to post it.

Point A:9.5m
Point B:6.5m
Point C:9.2m
Point D:0.50m
Point E:5.5m

a)What is its maximum speed?
b)With what speed does the roller coaster arrive at point E?
c)What constant braking would have to be applied at point E to stop the roller coaster within 5m.?

Analysis:
I finished part B. That was easy mgh=mgh+0.5mv^2 and the speed came out to be 8.8m/s

I have no idea how to do a and c. Before i begin. i just want to know, the more the energy the more the accelaration right. So at the heighst point of the track the roller coaster would have the max velocity. If this is right, then i know how to do part a as well. Use point A and point C to find the max veloctiy and point C.

Part c relates to the force of friction. I could use F=ma but i don't have the time so i am stuck here.

Hope you guys understand what i am saying.

Thanks
 
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ultimatejester said:
I have a exam tomorrow so i will greatly appreciate anyone who helps me on this one. The question has to be solved using law of conservation of energy.

Q: A roller coaster starts from rest at point A and weighs 1000kg with the ppl in it and goes through the designed track. The height of the track changes 5 times. There is a pic but i don't know how to post it.

Point A:9.5m
Point B:6.5m
Point C:9.2m
Point D:0.50m
Point E:5.5m

a)What is its maximum speed?
b)With what speed does the roller coaster arrive at point E?
c)What constant braking would have to be applied at point E to stop the roller coaster within 5m.?

Analysis:
I finished part B. That was easy mgh=mgh+0.5mv^2 and the speed came out to be 8.8m/s

I have no idea how to do a and c. Before i begin. i just want to know, the more the energy the more the accelaration right. So at the heighst point of the track the roller coaster would have the max velocity. If this is right, then i know how to do part a as well. Use point A and point C to find the max veloctiy and point C.

Part c relates to the force of friction. I could use F=ma but i don't have the time so i am stuck here.

Hope you guys understand what i am saying.

Thanks
Think of two glasses, with just enough water for one. The first glass is speed, and the other glass is height. You can pour water from the speed glass into the height glass, or vice versa, but you always have the same amount of water.

There will be the greatest possible amount of water in the speed glass when there is the least possible amount of water in the height glass. That is, the coaster will have the most kinetic energy when it has the least gravitational potential energy. So, to answer part A, you need to find out which part of the track is lowest. Whatever speed the coaster has there is the maximum speed it will achieve on the track.

You said, "the more energy the more acceleration right." This is not correct, kinetic energy describes only an object's velocity (and mass), it does not tell you how much it is accelerating.

The highest point of the track would have the least velocity.

Question C is worded ambiguously, they might be asking for the acceleration required to stop the car in five meters, or they might be asking for the amount of friction required to produce that acceleration. In either case, your first step should be to use your constant acceleration equations to find out how much acceleration would be required.
 
for reaching out for help with this question. In order to solve it using the law of conservation of energy, we need to consider the initial potential energy at point A and the final kinetic energy at point E. This is because according to the law of conservation of energy, energy cannot be created or destroyed, only transferred from one form to another.

a) To find the maximum speed, we need to find the point where all of the potential energy at point A is converted into kinetic energy. This would be at point D, where the roller coaster is at its lowest point. To solve for the maximum speed, we can use the equation mgh = 0.5mv^2, where m is the mass of the roller coaster (1000kg), g is the acceleration due to gravity (9.8m/s^2), h is the height difference between points A and D (9.5m-0.5m = 9m), and v is the maximum speed we are trying to find. Plugging in these values, we get v = √(2gh) = √(2*1000*9.8*9) = 42.4 m/s. So the maximum speed of the roller coaster is 42.4 m/s.

b) To find the speed at point E, we can use the same equation mgh = 0.5mv^2, but this time we are solving for v at point E. Plugging in the values for point E (m = 1000kg, g = 9.8m/s^2, h = 5.5m), we get v = √(2gh) = √(2*1000*9.8*5.5) = 35.3 m/s. So the roller coaster arrives at point E with a speed of 35.3 m/s.

c) To find the constant braking force required to stop the roller coaster within 5m, we can use the equation F = ma, where F is the force of friction, m is the mass of the roller coaster (1000kg), and a is the deceleration due to the braking force. We can find a by using the equation v^2 = u^2 + 2as, where u is the initial velocity (35.3 m/s) and s is the stopping distance (5m). Plugging in the values,
 

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