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: List given values

Mass $m=20\ \text{kg}$, kinetic friction $f_k=5\ \text{N}$, gravitational acceleration $g=9.8\ \text{m/s}^2$

Step2: Calculate normal force

Normal force equals weight: $N = m \times g = 20 \times 9.8 = 196\ \text{N}$

Step3: Solve for friction coefficient

Use $f_k = \mu_k N$, rearrange to $\mu_k = \frac{f_k}{N}$
$\mu_k = \frac{5}{196} \approx 0.02551$

Step4: Round to 3 decimals

Round $0.02551$ to 3 decimal places: $0.026$

Answer:

0.026