Determining Work Done by a Person on a Luggage

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SUMMARY

The discussion focuses on calculating the work done by a person pulling luggage up a ramp with a height of 5 meters and a distance of 20 meters, where the luggage weighs 10 kg. The formula used for work done while carrying a backpack is Wp = mgh, which is appropriate for vertical movement. However, the user questions the inclusion of kinetic friction and applied force in the equation for pulling luggage. The consensus is that the work done should be calculated using the work-energy theorem without subtracting friction from the force already accounted for in the weight of the luggage.

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  • Understanding of the work-energy theorem
  • Familiarity with gravitational potential energy calculations (Wp = mgh)
  • Knowledge of forces, including friction and applied force
  • Basic physics concepts related to work and energy
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  • Learn about calculating work done against friction
  • Explore the implications of different forces in work calculations
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SerenaMay
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I'm trying to determine the work done by a person as they pull a luggage up a ramp. The ramp has a height of 5 m and the distance the person walks up is 20 m. The weight of the bag is also 10 kg.

I am trying to compare the work done by pulling the luggage up a ramp to carrying an equally heavy backpack up a ramp on one's back. The formula I found and used to find the Work the person did when carrying the backpack was Wp = mgh (Alternatively, Fn(d*cos(theta)) = Fh*h = mgh ?).

I was wondering if the formula for the pulling would be the same and/or how would the kinetic friction and possibly applied force on the luggage trolley handle be added in the equation. Or if I am wrong in both the formulas, what would be a more correct way of doing it?

From what I am thinking of, Wp = mgh*(applied force - kinetic friction) since the kinetic friction is holding the luggage back a bit but it is still significantly less than the applied force of the person pulling on the handle.
 
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Hi SerenaMay and welcome to PF.

The most straightforward approach to this problem is the work-energy theorem. Have you studied it?
 
SerenaMay said:
Wp = mgh*(applied force - kinetic friction)
That equation cannot be right. You already have mg as a force, and you are multiplying by another force.
Also, we want the work done by the person. Why would you subtract the frictional force?

The person applies a force. How do you normally find the work done by a given force?
 

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