Question

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analyze this: a rightward force of 85.2 n is applied to a 3.14 - kg object to accelerate it to the right. the friction force between the object and the surface is 18.6 n. complete the diagram.
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units
force: n
mass: kg
acceln: m/s/s

Explanation:

Step1: Calculate \(F_{grav}\)

Use \(F_{grav}=mg\), where \(m = 3.14\space kg\) and \(g = 9.8\space m/s^{2}\).
\(F_{grav}=3.14\times9.8 = 30.772\space N\)

Step2: Determine \(F_{norm}\)

Since there is no vertical acceleration, \(F_{norm}=F_{grav}\). So \(F_{norm}=30.772\space N\)

Step3: Find \(F_{net}\)

\(F_{net}=F_{app}-F_{frict}\). Given \(F_{app}=85.2\space N\) and \(F_{frict}=18.6\space N\).
\(F_{net}=85.2 - 18.6=66.6\space N\)

Step4: Calculate acceleration \(a\)

Use \(F_{net}=ma\), so \(a=\frac{F_{net}}{m}\). Substitute \(F_{net}=66.6\space N\) and \(m = 3.14\space kg\).
\(a=\frac{66.6}{3.14}\approx21.21\space m/s^{2}\)

Answer:

\(F_{grav}=30.772\space N\), \(F_{norm}=30.772\space N\), \(F_{net}=66.6\space N\), \(a\approx21.21\space m/s^{2}\), \(m = 3.14\space kg\), \(F_{app}=85.2\space N\), \(F_{frict}=18.6\space N\)