Solving the Cart Friction Problem: Calculating Work Done by Force of Friction

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SUMMARY

The discussion focuses on calculating the work done by the force of friction on a 55 kg cart subjected to a horizontal force of 200 Newtons over a distance of 10 meters, while accelerating at 2 meters/second squared. The initial calculation of 900 Joules is incorrect because it fails to account for the net force acting on the cart. The correct approach requires using the net force, which is the sum of the applied force and the frictional force, to determine the work done by friction accurately.

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a horizontal force of 200 Newtons is applied to a 55 kg cart across a 10 meter surface. If the cart accelerates at 2 meters/second squared, then what is the work done by the force of friction as it acts to retard the motion of the cart ??

Work= FX
work= maX

55 kg x 2 x 10 = 1100 - 200 N = 900 Joules ??


is this correct ?
 
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williamx11373 said:
a horizontal force of 200 Newtons is applied to a 55 kg cart across a 10 meter surface. If the cart accelerates at 2 meters/second squared, then what is the work done by the force of friction as it acts to retard the motion of the cart ??

Work= FX
work= maX

55 kg x 2 x 10 = 1100 - 200 N = 900 Joules ??


is this correct ?
It is not correct. The "F" in F = ma is the net force, i.e. the sum of all the forces. In this case, it is the vector sum of the 200 N and the force of friction. What expression do you know that gives you the force of (kinetic) friction? Use that to calculate the work done by it.
 

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