Work done by friction (rolling resistance)

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SUMMARY

The discussion focuses on calculating the average frictional force acting on a carriage moving from point A to point B, where the carriage's mass is 4000 kg, the vertical drop is 110 m, and the maximum speed at B is 20 m/s. The gravitational potential energy (GPE) at A is calculated as 4,316,400 J, while the kinetic energy (KE) at B is 800,000 J, resulting in a work done of 3,516,400 J. However, it is concluded that the question lacks sufficient information to determine the average force, as the average force requires knowledge of momentum change over time, not just energy differences.

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Rumplestiltskin
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Homework Statement


The carriage and its passengers start at rest at A. At B, the bottom of the ride, the maximum speed of the carriage is 20ms-1. The vertical distance between A and B is 110m. The length of the track between A and B is 510m. The mass of the carriage and the passengers is 4000kg.

By considering the energy changes from A to B, determine the average frictional force acting on the carriage.

Homework Equations


The Attempt at a Solution



GPE at A = mgh = 4000 * 9.81 * 110 = 4316400 J
KE at B = 0.5 * 4000 * 202 = 800000 J
4316400 - 800000 = 3516400 J work done.
W = Fx. 3516400 = F * 510
F = 6900 N?

 
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Rumplestiltskin said:

Homework Statement


The carriage and its passengers start at rest at A. At B, the bottom of the ride, the maximum speed of the carriage is 20ms-1. The vertical distance between A and B is 110m. The length of the track between A and B is 510m. The mass of the carriage and the passengers is 4000kg.

By considering the energy changes from A to B, determine the average frictional force acting on the carriage.

Homework Equations


3. The Attempt at a Solution [/B]

GPE at A = mgh = 4000 * 9.81 * 110 = 4316400 J
KE at B = 0.5 * 4000 * 202 = 800000 J
4316400 - 800000 = 3516400 J work done.
W = Fx. 3516400 = F * 510
F = 6900 N?
Your arithmetic looks fine, but sadly the question is wrong. You do not have enough information to determine the average force.
Average force is ##\Delta p/\Delta t##, where p is momentum and t is time. This comes out the same as ##\Delta E/\Delta s## when the force is constant, but in not general.
See section 3 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/ for a fuller discussion.
 
haruspex said:
Your arithmetic looks fine, but sadly the question is wrong. You do not have enough information to determine the average force.
Average force is ##\Delta p/\Delta t##, where p is momentum and t is time. This comes out the same as ##\Delta E/\Delta s## when the force is constant, but in not general.
See section 3 of https://www.physicsforums.com/insights/frequently-made-errors-mechanics-forces/ for a fuller discussion.

Will be sure to tell my teacher. :P I guess this should suffice though. Thanks.
 

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