Question

analyze this: a rightward force of 339 n is applied to a 22.8 - kg object to accelerate it to the right. the coefficient of friction for the object and the surface is 0.547. complete the diagram.
tap on a field to enter or edit its value.
units
force: n
mass: kg
acceln: m/s/s
μ: no unit

Explanation:

Step1: Calculate \(F_{grav}\)

Using \(F_{grav}=mg\) where \(m = 22.8\space kg\) and \(g = 9.8\space m/s^{2}\)
\(F_{grav}=22.8\times9.8=223.44\space N\)

Step2: Determine \(F_{norm}\)

Since there is no vertical acceleration, \(F_{norm}=F_{grav}\)
\(F_{norm}=223.44\space N\)

Step3: Find \(F_{frict}\)

Using \(F_{frict}=\mu F_{norm}\), \(\mu = 0.547\) and \(F_{norm}=223.44\space N\)
\(F_{frict}=0.547\times223.44 = 122.22\space N\)

Step4: Calculate \(F_{net}\)

\(F_{net}=F_{app}-F_{frict}\), \(F_{app}=339\space N\) and \(F_{frict}=122.22\space N\)
\(F_{net}=339 - 122.22=216.78\space N\)

Step5: Find acceleration \(a\)

Using \(F_{net}=ma\), \(F_{net}=216.78\space N\) and \(m = 22.8\space kg\)
\(a=\frac{F_{net}}{m}=\frac{216.78}{22.8}=9.51\space m/s^{2}\)

Answer:

\(F_{norm}=223.44\space N\), \(F_{frict}=122.22\space N\), \(F_{app}=339\space N\), \(\mu = 0.547\), \(F_{grav}=223.44\space N\), \(m = 22.8\space kg\), \(a = 9.51\space m/s^{2}\), \(F_{net}=216.78\space N\)