Question

analyze this: a rightward force of 81.1 n is applied to a 5.5 - kg object to accelerate it to the right. the friction force between the object and the surface is 28.1 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 \(F_{grav}=mg\) (\(g = 9.8\ m/s^{2}\)), \(F_{grav}=5.5\times9.8=53.9\ N\)

Step2: Determine \(F_{norm}\)

Since there is no vertical acceleration (\(a_y = 0\)), by Newton's second law \(F_{net,y}=F_{norm}-F_{grav}=0\), so \(F_{norm}=F_{grav}=53.9\ N\)

Step3: Identify \(F_{frict}\)

Given \(F_{frict}=28.1\ N\)

Step4: Identify \(F_{app}\)

Given \(F_{app}=81.1\ N\)

Step5: Calculate \(F_{net}\)

Using \(F_{net}=F_{app}-F_{frict}\), \(F_{net}=81.1 - 28.1=53\ N\)

Step6: Calculate \(a\)

Using \(F_{net}=ma\), \(a=\frac{F_{net}}{m}=\frac{53}{5.5}\approx9.64\ m/s^{2}\)

Step7: Identify \(m\)

Given \(m = 5.5\ kg\)

Answer:

\(F_{norm}=53.9\ N\), \(F_{frict}=28.1\ N\), \(F_{app}=81.1\ N\), \(F_{grav}=53.9\ N\), \(m = 5.5\ kg\), \(a\approx9.64\ m/s^{2}\), \(F_{net}=53\ N\)