Question

A horizontal force of 200N is applied to a 55kg cart across a 10m level surface. If the cart accelerates at 2.0m/s^2, then what is the work done by the force of friction as it slows the motion of the cart?

Answer 1

1j=1newton*meter
force=mass*accel
200N=55x
200/55=3.636......
3.636...-2=1.636.....
1.636 is the deceleration resulting from friction hence the force of friction is 1.636*55=90newtons
90newtons*distance of 10 meters= 900 j of work done by friction

Answer 2

Work done by the force of friction 900 J and coefficient of friction Us=0.166

Step-by-step explanation:

∑
`F= m*a`

`F-F_(fr) = m*a`

`200 N - F_(fr) = 55 kg * 2 (m)/(s^(2) )  \\200 N - 110 N =  F_(fr)\\ F_(fr)= 90 N \\ F_(fr)= m*g*us  \\90N = 55 kg*9.8 (m)/(s^(2) ) *us  \\us= (90N)/(110N) = 0.166 \\W_(fr)= F_(fr)*d = 90N*10m = 900 J`