Question

1. find the accelerationof 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)

Explanation:

Step1: Calculate the normal force

The normal force \(N\) is the sum of the weight \(mg\) and the applied force \(F\).
\(N = mg+F\)
Given \(m = 2\space kg\), \(g = 10\space N/kg\), \(F=50\space N\)
\(N=(2\times10)+ 50=20 + 50=70\space N\)

Step2: Calculate the force of kinetic friction

The formula for the force of kinetic friction is \(f_k=\mu_kN\)
Given \(\mu_k = 0.5\), \(N = 70\space N\)
\(f_k=0.5\times70 = 35\space N\)

Step3: Apply Newton's second law \(F_{net}=ma\)

The net force \(F_{net}\) in the horizontal direction is \(F_{applied}-f_k\) (assuming the applied force is the only horizontal force, but here the problem might have a typo as it says "horizontal force of \(50\space N\)" which is vertical in the figure. But following the text: If we assume the \(50\space N\) is the horizontal force (contradicts figure, but text says so). Wait, no - re - check. Wait, the problem says "a horizontal force of \(50\space N\) acts on it on a rough surface". The normal force \(N=mg\) (if the \(50\space N\) is horizontal). Wait, there is a mis - match between figure and text. But text says "horizontal force of \(50\space N\)", so \(N = mg\) (because horizontal force doesn't contribute to normal force in standard case). Given \(m = 2\space kg\), \(g = 10\space N/kg\), \(N=2\times10=20\space N\)
\(f_k=\mu_kN=0.5\times20 = 10\space N\)
\(F_{net}=F - f_k\) (where \(F = 50\space N\) is horizontal), \(F_{net}=50-10=40\space N\)
By \(F_{net}=ma\), \(a=\frac{F_{net}}{m}\)
\(a=\frac{40}{2}=20\space m/s^{2}\)

Answer:

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