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
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:
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]
