Question

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?

Answer 1

Answer:
`68.75\ kg, 0.06`

Step-by-step explanation:

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

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

Friction opposes the motion of box

`\Rightarrow 73-f=m* 0.48\quad \ldots(i)\\\Rightarrow 84-f=m* 0.64\quad \ldots(ii)`

Subtract (i) from (ii)

`\Rightarrow 11=m(0.64-0.48)\\\Rightarrow m=68.75\ kg`

Therefore friction is

`\Rightarrow f=73-68.75* 0.48\\\Rightarrow f=73-33\\\Rightarrow f=40\ N`

Here, friction is kinetic friction which is given by

`\Rightarrow f=\mu_kmg\\\Rightarrow 40=\mu_k 68.75* 9.8\\\Rightarrow \mu_k=0.061`