How to Determine the Work Done by the Rope on a Sledge?

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Romain Nzebele
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Homework Statement


A sledge loaded with bricks has a total mass of 18.0 kg and is pulled at constant speed by a rope inclined at 20.0° above the horizontal. The sledge moves a distance of 20.0 m on a horizontal surface. The coefficient of kinetic friction between the sledge and surface is 0.500. How much work is done by the rope on the sledge?

Homework Equations


ΣF=ma, W=F.d

The Attempt at a Solution


I am not sure how to tackle this problem. Can I say that the work done on the sledge equal to zero because the acceleration of the sledge is zero and the force net is zero, or should I first calculate the tension of the rope and use this value to calculate the work done? I will appreciate any explanation. Thank you.
 
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Romain Nzebele said:
How much work is done by the rope on the sledge?
That's a bit confusing because work done by the rope and work done on the sledge are two different things. I would assume what's wanted is the work done by the rope.
I guess, in principle, the rope could have been doing work on two objects, and it would make sense to distinguish which work done component was wanted.
 
haruspex said:
That's a bit confusing because work done by the rope and work done on the sledge are two different things. I would assume what's wanted is the work done by the rope.
I guess, in principle, the rope could have been doing work on two objects, and it would make sense to distinguish which work done component was wanted.
If I understand well the concepts of work and force, the work done on the sledge will be equal to the force net on the sledge times distance, while the work done by the tension will be equal to the tension component along the displacement times the distance?
 
Romain Nzebele said:
If I understand well the concepts of work and force, the work done on the sledge will be equal to the force net on the sledge times distance, while the work done by the tension will be equal to the tension component along the displacement times the distance?
Yes.
 
Okay, thank you so very much.