Question

4.7 force of friction

1. find the acceleration of the object of mass m = 2 kg if a horizontal force of 50 n acts on it, on a rough surface where the coefficient of kinetic friction is 0.5, as shown in the figure. (take g = 10 n/kg)

m = 2 kg
f = 50 n
μ = 0.5

Explanation:

Step1: Calculate the normal force

On a horizontal surface, the normal force \(N\) equals the weight of the object. Using \(N = mg\), where \(m = 2\space kg\) and \(g=10\space N/kg\).
\(N=2\times10 = 20\space N\)

Step2: Calculate the frictional force

The formula for kinetic frictional force is \(F_f=\mu N\). Given \(\mu = 0.5\) and \(N = 20\space N\).
\(F_f=0.5\times20=10\space N\)

Step3: Apply Newton's second law

Newton's second law is \(F_{net}=ma\), and \(F_{net}=F - F_f\). Here \(F = 50\space N\) and \(F_f = 10\space N\), \(m = 2\space kg\).
\(F - F_f=ma\)
\(50 - 10=2a\)
\(40 = 2a\)
\(a=\frac{40}{2}=20\space m/s^{2}\)

Answer:

\(20\space m/s^{2}\)