Question

multiple choice 1 point
a box with a mass of 15.0 kg is being pushed on a level surface with a horizontally applied force. it just starts to slide when 40.0 n is applied and will continue to slide when 35.0 n is applied to the box.
how much is the normal force (fn) acting on the box. (remember that direction is part of force.).
-150 n 150 n -35.0 n 35.0 n -40.0 n 40.0 n -75.0 n 75.0 n -150. n 150. n none of these
friction problems
\\( \vec{f}_s = \mu_s \vec{f}_n \\) \\( \vec{f}_k = \mu_k \vec{f}_n \\) \\( \vec{f}_g = m\vec{g} \\)

Explanation:

Step1: Analyze forces in vertical - direction

On a level surface, the normal force \(F_N\) balances the weight of the box. The weight \(F_g=mg\), but we are not asked to calculate the weight. In the vertical direction, there are no vertical - applied forces. The normal force is perpendicular to the surface. When the box is on a level surface and no vertical forces (other than gravity and the normal force) act on it, the normal force \(F_N\) has no component in the horizontal direction. The horizontal forces are the applied force and the friction force. The normal force is a contact force perpendicular to the surface of contact between the box and the surface.

Step2: Determine the nature of the normal force

The normal force \(F_N\) acts perpendicular to the surface. Its direction is away from the surface and towards the box. In the context of the problem, we are interested in the magnitude and direction of the normal force. Since the box is on a level surface and there are no vertical - applied forces, the normal force \(F_N\) has a magnitude equal to the weight of the box in the static and kinetic cases. However, the question asks for the normal force in terms of direction. The normal force on the box acts in the upward direction. In vector notation, if we consider the upward direction as positive and the downward direction as negative, and assume the surface is horizontal, the normal force \(F_N\) is positive. The normal force is not related to the horizontal - applied forces \(40.0\ N\) and \(35.0\ N\) in terms of direction or magnitude in the sense that they are in different directions (horizontal and vertical). The normal force magnitude \(F_N = mg\) (where \(m = 15.0\ kg\) and \(g=9.8\ m/s^2\)), but we are not calculating the magnitude here. The direction of the normal force on the box is upward. If we consider the coordinate system where the upward - direction is positive, the normal force \(F_N\) has a positive value in the vertical direction.

Answer:

150 N (assuming \(g = 9.8\ m/s^2\), \(m = 15.0\ kg\), \(F_N=mg=15\times10 = 150\ N\) approximately, and the positive value indicates the upward direction which is the direction of the normal force on a level - surface situation without any vertical applied forces. The closest positive value to the calculated \(mg\) among the options is 150 N)