Question

a boy pushes a 5000kg box along a floor with a force of 2000N.if the velocity of the box is uniform.What is the coefficient of friction between the box and the floor

Answer 1

Since the speed is constant (zero acceleration) the friction force is equal to the pushing force by the second Newton's law:

$$F_{fr} = 2000N$$

On the other hand, the friction force is given as follows:

$$F_{fr} = \mu N$$

where $\mu$ is the coefficient of friction, and $N$ is the normal reation force, which is equal to the weight of the box, since the box does not move in vertical direction. Thus, obtain:

$$F_{fr} = \mu mg$$

where $m =5000kg$ is the mass of the box, and $g = 9.8m/s^2$ is the gravitational acceleration. Finally, have:

$$\mu = \dfrac{F_{fr}}{mg} = \dfrac{2000N}{5000kg\cdot 9.8m/s^2} \approx 0.04$$

Answer. 0.04.