Mary applies a force of 73 N to push a box with an acceleration of 0.59 m/s 2 .

Question

Mary applies a force of 73 N to push a box with an acceleration of 0.59 m/s2. When she increases the pushing force to 82 N, the box's acceleration changes to 0.81 m/s2. There is a constant friction force present between the floor and the box.     (a) What is the mass of the box?
1    kg

    (b) What is the coefficient of kinetic friction between the floor and the box?
2     (a) What is the mass of the box?
1    kg

    (b) What is the coefficient of kinetic friction between the floor and the box?
2

Explanation / Answer

let , the constant frictional force be f N

then,

73-f=m*0.59 ---(1)

82-f=m*0.81 ---(2)

solving 1 and 2 ,we get

mass,m=450/11=40.91 kg

frictional force,f= 1075/22=48.86 N

Normal reaction,N=mg=40.91*9.81=401.33 N

f=u*N

=>u=48.86/401.33=0.122

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