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analyze this: a rightward force is applied to a 6.52 - kg object to accelerate it to the right at 12.1 m/s/s. the friction force between the object and the surface is 25.8 n. complete the diagram.
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force: n
mass: kg
acceln: m/s/s
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wbfsmarto82@student.wbsd.org (replay - no tracking)
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master
problem 3 of 3
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Explanation:

Step1: Calculate net force

According to Newton's second law \(F_{net}=ma\). Given \(m = 6.52\space kg\) and \(a=12.1\space m/s^{2}\), then \(F_{net}=6.52\times12.1 = 78.9\space N\) (right - ward).

Step2: Relate applied force, friction force and net force

We know that \(F_{net}=F_{app}-F_{frict}\). Rearranging for \(F_{app}\), we get \(F_{app}=F_{net}+F_{frict}\). Given \(F_{frict} = 25.8\space N\) and \(F_{net}=78.9\space N\), then \(F_{app}=78.9 + 25.8=104.7\space N\)

Step3: Calculate gravitational and normal forces

The gravitational force \(F_{grav}=mg\), where \(g = 9.8\space m/s^{2}\) and \(m = 6.52\space kg\). So \(F_{grav}=6.52\times9.8=63.9\space N\). In the vertical direction, since there is no vertical acceleration (\(a_y = 0\)), by Newton's first law \(F_{norm}=F_{grav}\), so \(F_{norm}=63.9\space N\)

Answer:

\(F_{norm}=63.9\space N\), \(F_{app}=104.7\space N\), \(F_{grav}=63.9\space N\), \(m = 6.52\space kg\), \(a = 12.1\space m/s^{2}\), \(F_{net}=78.9\space N\), \(F_{frict}=25.8\space N\)