Question

you are pushing a 10 kg box on a horizontal surface with a force of 45 n and the box moves with constant speed. what is the coefficient of kinetic friction between the box and the surface?
a 4.5
b 0.46
c 45
d since it is moving there is no friction

Explanation:

Step1: Analyze the forces

Since the box moves with constant speed, the net force is zero. The applied force \(F = 45\ N\) is equal to the kinetic - friction force \(F_f\). So \(F_f=F = 45\ N\).
The normal force \(N\) on the box on a horizontal surface is \(N = mg\), where \(m = 10\ kg\) and \(g=9.8\ m/s^{2}\). Then \(N=10\times9.8 = 98\ N\).

Step2: Use the formula for kinetic - friction

The formula for kinetic - friction is \(F_f=\mu_kN\), where \(\mu_k\) is the coefficient of kinetic friction.
We know \(F_f = 45\ N\) and \(N = 98\ N\). Rearranging the formula for \(\mu_k\) gives \(\mu_k=\frac{F_f}{N}\).
Substitute the values: \(\mu_k=\frac{45}{98}\approx0.46\)

Answer:

B. 0.46