Question

Mary applies a force of 71 N to push a box with an acceleration of 0.57 m/s2. When she increases the pushing force to 82 N, the box's acceleration changes to 0.80 m/s2. There is a constant friction force present between the floor and the box.

(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:

(a). The mass of the box is 47.8 kg.

(b). The coefficient of kinetic friction between the floor and the box is 0.093.

Explanation:

Given that,

Force = 71 N

Acceleration = 0.57 m/s²

Pushing force = 82 N

Change acceleration = 0.80 m/s²

(a). We need to calculate the mass of the box

Using balance equation

$N-f=ma$

Put the value in the equation

$71-f=m\times0.57$....(I)

$82-f=m\times0.80$....(II)

From equation (I) and (II)

$82-71=m(0.80-0.57)$

$11=m\times0.23$

$m=\dfrac{11}{0.23}$

$m=47.8\ kg$

The mass of the box is 47.8 kg.

(b). We need to calculate the friction force

Now, put the value of m in equation (I)

$71-f=47.8\times0.57$

$-f=47.8\times0.57-71$

$f=43.75\ N$

We need to calculate the coefficient of kinetic friction between the floor and the box

Using friction force

$f=\mu mg$

$\mu=\dfrac{f}{mg}$

$\mu=\dfrac{43.75}{47.8\times9.8}$

$\mu=0.093$

The coefficient of kinetic friction between the floor and the box is 0.093.

Hence, (a). The mass of the box is 47.8 kg.

(b). The coefficient of kinetic friction between the floor and the box is 0.093.