Question

1. the coefficient of kinetic friction acting between the bottom of a 10.0 kg box and the floor is 0.30. if the box is pulled to the right by a force of 75.0 n, determine the acceleration of the box. 5. calculate the acceleration of the box below. assume a 25.0 n force of friction

Explanation:

Step1: Calculate the kinetic friction force

The formula for kinetic friction force is \(F_f=\mu_k F_N\). Since the box is on a horizontal floor, \(F_N = mg\) (where \(m = 10.0\space kg\) and \(g= 9.8\space m/s^2\)). So \(F_N=10.0\times9.8 = 98\space N\). Then \(F_f=\mu_k F_N=0.30\times98 = 29.4\space N\)

Step2: Apply Newton's second law

Newton's second law is \(F_{net}=ma\), and \(F_{net}=F - F_f\) (where \(F = 75.0\space N\) is the applied force). So \(F_{net}=75.0- 29.4=45.6\space N\). Then \(a=\frac{F_{net}}{m}\), substituting \(m = 10.0\space kg\) and \(F_{net}=45.6\space N\), we get \(a=\frac{45.6}{10.0}=4.56\space m/s^2\)

Answer:

\(4.56\space m/s^2\)