Question

a box experiencing a gravitational force of 600 n is being pulled to the right with force of 250 n. a 25 n frictional force acts on the box as it moves to the right. what is the net force in the y-direction?
250 n
25 n
0 n
225 n

Explanation:

Step1: Identify forces in y - direction

In the vertical (y - direction), the gravitational force ($F_g = 600\ N$) acts downward, and the normal force ($F_N$) acts upward. For an object on a horizontal surface (implied here as the box is moving horizontally), the normal force and gravitational force are equal in magnitude and opposite in direction when the object is in equilibrium vertically (not accelerating up or down).

Step2: Calculate net force in y - direction

The net force in the y - direction is given by the sum of the forces in the y - direction. Let the upward direction be positive and downward be negative. So $F_{net,y}=F_N - F_g$. Since $F_N = F_g$ (from vertical equilibrium, as there is no acceleration in y - direction), then $F_{net,y}=F_g - F_g=0\ N$.

Answer:

0 N (corresponding to the option with 0 N)