Work and Energy

Question

The figure below shows a roller coaster. The 500 kg carriage is pulled up the first incline from ground level at A to the first peak at B. From here it then rolls freely from rest under the influence of gravity.

 

a) Describe the energy transformations involved when the cart travels from B to C.

i) If the track is frictionless:
Moving from A to B: __________________________
Moving from B to C: __________________________
Moving from C to D: __________________________
ii) If we included friction:
Moving from A to B: __________________________
Moving from B to C: __________________________
Moving from C to D: __________________________

 

b) If no friction forces are present, how fast will the cart be moving at location C?

c) Upon reaching location C the card continues to roll freely to location D. How much gravitational potential energy does the cart gain from C to D?

d) Based on your answer above, how much kinetic energy does the cart lose from C to D? Where did the kinetic energy go?

e) What is the final speed of the cart at location D?

f) In reality, we will always have friction. The final speed of the cart at location D is found to be only 13 m/s. What is the total amount of mechanical energy (both potential and kinetic) at location D?

g) How much energy was “lost” overall from location B to D?

h) Efficiency is a percentage that describes how much energy was conserved as “useful” energy between B and D. What is the efficiency of this coaster?

 

 

 

 

 

 

 

 

 

Answer 1

If the track is assumed to be frictionless, then the energy change is from mechanical energy or in this case mechanical work due to the pulling force to potential gravitational energy from A to B.  From B to C, I can assume that the potential energy will be converted to kinetic energy as it must be going downhill, but unless it goes as low as A or lower, then not all potential energy will be transformed to kinetic.  Again, assuming the car then goes back uphill partially from C to D, then some of the kinetic energy is converted back to potential energy.

 

If we include friction into the problem, then part of the energy will be "lost" do to work from friction which would be included in all stages.

 

For part b, the potential energy (mass*gravity*height)=1/2*mass*velocity^2, where the height is the difference in elevation between b and c.

 

For parts c and d, mass*gravity*height, where this height is now the difference in elevation between c and d will give the energy gained back to potential or therefore lost from kinetic.

 

For part e, the potential energy difference between parts b and c can be used to set equal again to 1/2*mass*velocity^2 to solve for velocity.

 

For part f, we know gravitational potential energy is equal to mass*gravity*height and kinetic energy is equal to 1/2*mass*velocity^2.  The sum of these two is the total mechanical energy at D

 

For part g, the gravitational potential energy was equal to mass*gravity*height at that point, so this was the total mechanical energy available at the start of the first hill where the roller coaster was initially able to be viewed as at rest for a brief moment.  subtract part f from this original potential energy and you have what was lost.

 

For part h, efficiency would be to divide the original potential energy at point b found in part g by the total mechanical energy found in part f and multiplying be 100 to obtain the efficiency as a percentage.