- #1
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.
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.