Question

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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.
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 \(F_{grav}=mg\) (\(g = 9.8\ m/s^{2}\)), \(F_{grav}=6.52\times9.8 = 63.896\ N\)

Step2: Calculate \(F_{norm}\)

Since there is no vertical acceleration, \(F_{norm}=F_{grav}\), so \(F_{norm}=63.896\ N\)

Step3: Calculate \(F_{net}\)

Using \(F_{net}=ma\), \(F_{net}=6.52\times12.1 = 78.892\ N\)

Step4: Calculate \(F_{app}\)

Using \(F_{net}=F_{app}-F_{frict}\), then \(F_{app}=F_{net}+F_{frict}\), \(F_{app}=78.892 + 25.8=104.692\ N\)

Answer:

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