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

[williamx11373]

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 ?

 

[kuruman]

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.