Mary applies a force of 73 N to push a box with an acceleration of 0.48 m/s^2. When she increases the pushing force to 84 N, the box's acceleration changes to 0.64 m/s^2. There is a constant friction force present between the floor and the box.

Required:
a. What is the mass of the box?
b. What is the coefficient of kinetic friction between the floor and the box?

Respuesta :

Answer: [tex]68.75\ kg, 0.06[/tex]

Explanation:

Mary applies a force of 73 N to create an acceleration of [tex]0.48\ m/s^2[/tex]

When She increases force to 84 N, it creates an acceleration of [tex]0.64\ m/s^2[/tex]

Friction opposes the motion of box

[tex]\Rightarrow 73-f=m\times 0.48\quad \ldots(i)\\\Rightarrow 84-f=m\times 0.64\quad \ldots(ii)[/tex]

Subtract (i) from (ii)

[tex]\Rightarrow 11=m(0.64-0.48)\\\Rightarrow m=68.75\ kg[/tex]

Therefore friction is

[tex]\Rightarrow f=73-68.75\times 0.48\\\Rightarrow f=73-33\\\Rightarrow f=40\ N[/tex]

Here, friction is kinetic friction which is given by

[tex]\Rightarrow f=\mu_kmg\\\Rightarrow 40=\mu_k 68.75\times 9.8\\\Rightarrow \mu_k=0.061[/tex]

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