Question

1. a 20 kg crate of mangos experiences 5 n of frictional force when pushed at a constant speed. what is the coefficient of kinetic friction between these two surfaces? round to 3 decimal places.

Explanation:

Step1: Recall kinetic friction formula

The formula for kinetic friction is $F_k = \mu_k F_N$, where $F_k$ is kinetic friction force, $\mu_k$ is coefficient of kinetic friction, and $F_N$ is normal force.

Step2: Calculate normal force

For a horizontal surface, normal force equals weight: $F_N = mg$.
$m=20\ \text{kg}$, $g=9.8\ \text{m/s}^2$, so $F_N = 20 \times 9.8 = 196\ \text{N}$.

Step3: Solve for $\mu_k$

Rearrange formula: $\mu_k = \frac{F_k}{F_N}$.
Substitute $F_k=5\ \text{N}$, $F_N=196\ \text{N}$: $\mu_k = \frac{5}{196}$

Step4: Compute and round value

$\mu_k \approx 0.02551$, rounded to 3 decimals is $0.026$.

Answer:

0.026