A 5 kg box is sitting on a table. The coefficient of static friction between the box and the table is 0.4. What is the minimum horizontal force needed to move the box?

Step 1 ：Step 1: Identify the forces acting on the box. The weight of the box is given by \(mg\), where \(m = 5 \, kg\) and \(g = 9.8 \, m/s^2\). The force of static friction is given by \(\mu_s F_N\), where \(\mu_s = 0.4\) is the coefficient of static friction and \(F_N\) is the normal force. Since the box is not moving vertically, the normal force equals the weight of the box.

Step 2 ：Step 2: Calculate the weight of the box. Substitute \(m = 5 \, kg\) and \(g = 9.8 \, m/s^2\) into the equation \(mg\) to get \(F_N = 5 \, kg \cdot 9.8 \, m/s^2 = 49 \, N\).

Step 3 ：Step 3: Calculate the force of static friction. Substitute \(\mu_s = 0.4\) and \(F_N = 49 \, N\) into the equation \(\mu_s F_N\) to get \(F_s = 0.4 \cdot 49 \, N = 19.6 \, N\).

Step 4 ：Step 4: The minimum horizontal force needed to move the box equals the force of static friction, which is 19.6 N.