Question

1. if a 5 kg mass experiences a 2 n force of friction when sliding, what is the coefficient of kinetic friction between these two surfaces? round to 2 decimal places.

Explanation:

Step1: Recall friction force formula

The kinetic friction force is given by $F_f = \mu_k F_N$, where $F_N$ is the normal force. For a mass on a horizontal surface, $F_N = mg$.

Step2: Calculate normal force

$F_N = mg = 5\ \text{kg} \times 9.8\ \text{m/s}^2 = 49\ \text{N}$

Step3: Solve for coefficient $\mu_k$

Rearrange the formula: $\mu_k = \frac{F_f}{F_N}$
Substitute values: $\mu_k = \frac{2\ \text{N}}{49\ \text{N}}$

Step4: Compute and round result

$\mu_k \approx 0.0408 \approx 0.04$

Answer:

0.04