Question

analyze this: a rightward force is applied to a 6.11 - kg object to accelerate it to the right at 15.1 m/s/s. the friction force between the object and the surface is 28.6 n. complete the diagram. tap on a field to enter or edit its value. units force: n mass: kg acceln: m/s/s

Explanation:

Step1: Calculate \(F_{grav}\)

Using the formula \(F_{grav}=mg\) where \(m = 6.11\space kg\) and \(g = 9.8\space m/s^{2}\).
\(F_{grav}=6.11\times9.8=59.878\space N\)

Step2: Determine \(F_{norm}\)

Since there is no vertical acceleration (\(a_y = 0\)), by Newton's second law \(F_{net,y}=F_{norm}-F_{grav}=ma_y = 0\). So \(F_{norm}=F_{grav}\).
\(F_{norm}=59.878\space N\)

Step3: Calculate \(F_{net}\)

Using Newton's second law \(F_{net}=ma\) with \(m = 6.11\space kg\) and \(a=15.1\space m/s^{2}\).
\(F_{net}=6.11\times15.1 = 92.261\space N\)

Step4: Find \(F_{app}\)

Using \(F_{net}=F_{app}-F_{frict}\). Rearranging for \(F_{app}\), we get \(F_{app}=F_{net}+F_{frict}\). Given \(F_{frict}=28.6\space N\) and \(F_{net}=92.261\space N\).
\(F_{app}=92.261 + 28.6=120.861\space N\)

Answer:

\(m = 6.11\space kg\)
\(a = 15.1\space m/s^{2}\)
\(F_{grav}=59.878\space N\)
\(F_{norm}=59.878\space N\)
\(F_{net}=92.261\space N\)
\(F_{app}=120.861\space N\)
\(F_{frict}=28.6\space N\)