Two horizontal forces, F1 and F2, are acting on a box, but only F1 is shown in the drawing. F2 can point either to the right or to the left. The box moves only along the x axis. There is no friction between the box and the surface. Suppose that = +6.2 N and the mass of the box is 3.3 kg. Find the magnitude and direction of when the acceleration of the box is (a) +5.5 m/s2, (b) -5.5 m/s2, and (c) 0 m /s2.

BOX----->F1--+x

Fnet= F1 + F2

Fnet=mass * acceleration

+6.2+ F2= (3.3)a

A) F2= (3.3)(+5.5) - 6.2
B) F2= (3.3)(-5.5) - 6.2
C) F2= 0 - 6.2

6.2

To find the magnitude and direction of the second force (F2) acting on the box, we can use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration. Mathematically, this can be represented as:

Net force (F_net) = Mass (m) * Acceleration (a)

Given:
Force exerted by F1 (F1) = +6.2 N
Mass of the box (m) = 3.3 kg

Let's calculate the net force (F_net) required for each case:

(a) When the acceleration of the box is +5.5 m/s²:
Since there is no friction and only F1 is acting, we can assume F_net = F1.
F_net = F1 = 6.2 N (to the right)

(b) When the acceleration of the box is -5.5 m/s²:
Again, since there is no friction and only F1 is acting, we can assume F_net = F1.
F_net = F1 = 6.2 N (to the right)

(c) When the acceleration of the box is 0 m/s²:
In this case, the net force acting on the box must be zero to have zero acceleration. Since F1 is the only force acting, we can assume F_net = F1 - F2, where F2 is the force we need to find.
F_net = F1 - F2 = 0
F2 = F1 = 6.2 N (to the right)

Therefore, the magnitude of the force F2 acting on the box is 6.2 N, and its direction is to the right (in the same direction as force F1) in all three cases.

To find the magnitude and direction of F2 when the acceleration of the box is given, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

Let's start with the equation for the net force acting on the box:

Net Force = F1 + F2

Since there is no friction between the box and the surface, the only forces acting on the box are F1 and F2.

Now, let's consider the given situations:

(a) When the acceleration of the box is +5.5 m/s^2:
We can substitute the known values into the net force equation and solve for F2:

Net Force = (mass of the box) * (acceleration)
F1 + F2 = (3.3 kg) * (5.5 m/s^2)
F1 + F2 = 18.15 N

Since F1 is given as +6.2 N, we can subtract it from the net force to find F2:

F2 = Net Force - F1 = 18.15 N - 6.2 N = 11.95 N

The magnitude of F2 when the acceleration is +5.5 m/s^2 is 11.95 N. The direction of F2 can be determined from the drawing, where it is mentioned that F2 can point either to the right or to the left.

(b) When the acceleration of the box is -5.5 m/s^2:
Using the same approach, we have:

Net Force = F1 + F2 = (3.3 kg) * (-5.5 m/s^2)
F1 + F2 = -18.15 N

Solving for F2:

F2 = Net Force - F1 = -18.15 N - 6.2 N = -24.35 N

The magnitude of F2 when the acceleration is -5.5 m/s^2 is 24.35 N. The negative sign indicates that the force is in the opposite direction to F1, so F2 is to the left.

(c) When the acceleration of the box is 0 m/s^2:
In this case, there is no net force acting on the box, as it is not accelerating. Therefore:

Net Force = F1 + F2 = 0
F2 = -F1

Substituting the given value of F1:

F2 = -6.2 N

The magnitude of F2 when the acceleration is 0 m/s^2 is 6.2 N. The negative sign indicates that the force is in the opposite direction to F1, so F2 is to the left.