Question

1. 2 points a box of mass of 45.0 kg is being pushed on a level surface with a horizontally applied force. the coefficient of static friction (μs) is 0.500 and the coefficient of kinetic friction (μk) is 0.400. how much is the normal force (fn) acting on the box? (remember that direction is part of force). type your answer— known coefficients fs=μsfn fk=μkfn fg=mg

Explanation:

Step1: Identify the forces acting on the box vertically

In the vertical - direction, for a box on a level surface, the normal force \(F_N\) balances the weight of the box. The weight of the box is given by \(F_g=mg\), where \(m = 45.0\ kg\) and \(g = 9.8\ m/s^2\).

Step2: Calculate the normal force

Using the formula \(F_N=mg\), we substitute \(m = 45.0\ kg\) and \(g = 9.8\ m/s^2\). So \(F_N=45.0\times9.8 = 441\ N\).

Answer:

441 N