Question

part a
a 14 kg box must be slid across the floor. if the coefficient of static friction between the box and floor is 0.37, what is the minimum force needed to start the box moving from rest?
138 n
36 n
51 n
71 n
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Explanation:

Step1: Calculate the normal force

The normal force \(N\) on the box is equal to its weight \(mg\) (where \(m = 14\space kg\) and \(g=9.8\space m/s^{2}\)). So \(N = mg=14\times9.8 = 137.2\space N\)

Step2: Calculate the maximum static - friction force

The formula for the maximum static - friction force \(f_s=\mu_sN\), where \(\mu_s = 0.37\) and \(N = 137.2\space N\). Then \(f_s=0.37\times137.2\approx50.764\space N\approx51\space N\)

Answer:

C. \(51\space N\)