Question

question 19
1 pts
a box has a mass of 8 kilograms and sits on a rough, flat floor. the box is pushed to the right by an applied force ( f_a ) and feels a kinetic friction force to the left. the object has a coefficient of kinetic friction equal to 0.2 and accelerates to the right with an acceleration of ( 12 mathrm{~m} / mathrm{s}^2 ). determine the magnitude of the applied force in newtons.
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Explanation:

Step1: Calculate the normal force

On a flat surface, the normal force \(N\) equals the weight of the box. Using \(N = mg\) (where \(m = 8\space kg\) and \(g= 9.8\space m/s^{2}\)), we have \(N=8\times9.8 = 78.4\space N\).

Step2: Calculate the kinetic - friction force

The formula for kinetic - friction force is \(F_{f}=\mu_{k}N\). Given \(\mu_{k}=0.2\) and \(N = 78.4\space N\), then \(F_{f}=0.2\times78.4=15.68\space N\).

Step3: Apply Newton's second law

Newton's second law is \(F_{net}=ma\). The net force \(F_{net}=F_{A}-F_{f}\). We know \(a = 12\space m/s^{2}\) and \(m = 8\space kg\), so \(F_{A}-F_{f}=ma\). Rearranging for \(F_{A}\), we get \(F_{A}=ma + F_{f}\).
Substitute \(m = 8\space kg\), \(a=12\space m/s^{2}\), and \(F_{f}=15.68\space N\) into the equation: \(F_{A}=8\times12+15.68\).
First, calculate \(8\times12 = 96\), then \(F_{A}=96 + 15.68=111.68\space N\).

Answer:

\(111.68\)