Question

the free-body diagram below represents the forces acting on an object. how can the net force be determined? f_n = 123 n, f_r = -90 n, f_a = 150 n, f_g = -123 n. options: combine the horizontal forces only; combine the vertical forces only; combine the vertical and horizontal forces; the net force can not be determined. based on the velocity vs. time graph below, how can the acceleration between 0 and 5 seconds be determined? (acceleration = change in velocity / change in time)

Explanation:

To determine the net force, we analyze the vertical and horizontal forces. Vertically, \( F_N = 123\,\text{N} \) (upward) and \( F_g = - 123\,\text{N} \) (downward), so their vector sum is \( 123 + (-123)=0\,\text{N} \). Horizontally, \( F_r=-90\,\text{N} \) (left) and \( F_A = 150\,\text{N} \) (right). The net force is the vector sum of all forces, but since vertical forces cancel, we only need to combine horizontal forces? Wait, no—net force is the vector sum of all forces (vertical and horizontal). Wait, but in this case, vertical forces are equal and opposite (equilibrium), so the net force comes from horizontal forces. Wait, the options: "combine the horizontal forces only"—but actually, net force is the vector sum of all forces. But let's check the forces: vertical forces: \( F_N = 123\,\text{N} \) (positive y - direction) and \( F_g=-123\,\text{N} \) (negative y - direction). Their sum is \( 123+( - 123)=0\,\text{N} \). Horizontal forces: \( F_r = - 90\,\text{N} \) (negative x - direction) and \( F_A=150\,\text{N} \) (positive x - direction). The net force is the sum of all forces, but since vertical forces sum to zero, the net force is equal to the sum of horizontal forces. But the question is "how can the NET force be determined?". The vertical forces cancel out (their vector sum is zero), so the net force is just the sum of the horizontal forces. Wait, but the option "combine the horizontal forces only"—but technically, we should combine all forces, but since vertical forces cancel, combining horizontal forces gives the net force. Alternatively, maybe the intended logic is: vertical forces are balanced (equal magnitude, opposite direction), so their net is zero. So to find the overall net force, we only need to combine the horizontal forces (since vertical net is zero). So the correct option is "combine the horizontal forces only"? Wait, no—wait, net force is the vector sum of all forces. Let's calculate:

Net force in y - direction: \( F_{net,y}=F_N + F_g=123+( - 123) = 0\,\text{N} \)

Net force in x - direction: \( F_{net,x}=F_A+F_r=150+( - 90)=60\,\text{N} \)

Overall net force: \( \vec{F}_{net}=\vec{F}_{net,x}+\vec{F}_{net,y}=60\,\text{N}\hat{i}+0\,\text{N}\hat{j} \), so the net force is determined by combining horizontal forces (since vertical forces cancel). So the correct option is "combine the horizontal forces only"? Wait, but the option "combine the vertical and horizontal forces"—but vertical forces sum to zero. So if we combine vertical and horizontal, we get the same as combining horizontal (since vertical is zero). But the options: let's re - evaluate. The forces are:

Vertical: normal force (up) and gravity (down), equal in magnitude, opposite in direction. So their vector sum is zero.

Horizontal: applied force (right) and friction (left), different magnitudes. So the net force is the vector sum of all four forces, but since vertical forces cancel, the net force is just the sum of the horizontal forces. So to determine the net force, we can combine the horizontal forces only (because vertical forces cancel out). So the correct option is "combine the horizontal forces only"? Wait, no—the net force is the sum of all forces. But since vertical forces sum to zero, the net force is equal to the sum of horizontal forces. So the way to determine the net force is to combine the horizontal forces only (because vertical forces don't contribute to the net force as they cancel).

Answer:

A. combine the horizontal forces only (assuming the first option is A, but in the given options, the first option is "combine the horizontal forces only")