Calculating Work Done by Friction on a 55kg Cart with 200N Horizontal Force

[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 ??

 

[Andrew Mason]

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 ??

Oddly enough your answer is correct. But I have no idea what you are doing.

Do a free body diagram. What are the forces acting? What do these necessarily sum to? That will allow you to determine the force of friction. Over what distance does the force of friction act? That force x the distance over which it acts gives you the work done by friction.

AM