Question

You are moving a 50.0 kg crate across a flat, horizontal surface at constant velocity with a force of 70.0 N. What is the coefficient of kinetic friction between the crate and the floor?

Answer 1

To determine the coefficient of kinetic friction between the crate and the floor, we can use the equation:

$F_{\text{friction}}={\mu }_k\cdot F_{\text{normal}}$

where: $F_{\text{friction}}$ is the force of kinetic friction, ${\mu }_k$ is the coefficient of kinetic friction, and $F_{\text{normal}}$ is the normal force.

In this case, the crate is moving at a constant velocity, which means the force of friction is equal to the applied force. Therefore, we have:

$F_{\text{friction}}=70.0\text{N}$

The normal force ($F_{\text{normal}}$) is equal to the weight of the crate ($m\cdot g$), where $m$ is the mass of the crate and $g$ is the acceleration due to gravity (approximately 9.8 m/s²):

Fnormal=m⋅g=50.0 kg×9.8 m/s2F_{ ext{normal}} = m cdot g = 50.0 , ext{kg} imes 9.8 , ext{m/s}^2F<span cla