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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 :

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
davidlcastiblanco

Answer:

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

Explanation:

∑[tex]F= m*a[/tex]

[tex]F-F_{fr} = m*a[/tex]

[tex]200 N - F_{fr} = 55 kg * 2 \frac{m}{s^{2} }  \\200 N - 110 N =  F_{fr}\\ F_{fr}= 90 N \\ F_{fr}= m*g*us  \\90N = 55 kg*9.8 \frac{m}{s^{2} } *us  \\us= \frac{90N}{110N} = 0.166 \\W_{fr}= F_{fr}*d = 90N*10m = 900 J[/tex]

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