Question

3 a wooden crate falls out of the back of a contractors truck. when it lands on the concrete road, it begins sliding and friction causes it to slow down. if the crate has a mass of 23 kg and the force of friction is 128 n, what is the coefficient of kinetic friction between the wood crate and the concrete road?

μₖ = 0.62

μₖ = 0.16

μₖ = 0.02

μₖ = 0.54

Explanation:

Step1: Calculate the normal force

The normal force \(N\) on a horizontal surface is equal to the weight of the object. Using the formula \(N = mg\), where \(m = 21\space kg\) and \(g=9.8\space m/s^{2}\).
\(N=21\times9.8 = 205.8\space N\)

Step2: Use the formula for kinetic friction

The formula for kinetic friction is \(F_{k}=\mu_{k}N\). We know \(F_{k} = 128\space N\) and \(N = 205.8\space N\). Rearranging for \(\mu_{k}\), we get \(\mu_{k}=\frac{F_{k}}{N}\)
\(\mu_{k}=\frac{128}{205.8}\approx0.62\)

Answer:

\(\mu_{k} = 0.62\)