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 F1-> = +8.0 N and the mass of the box is 4.7 kg. Find the magnitude and direction of F2-> when the acceleration of the box is (a) +6.7 m/s2, (b) -6.7 m/s2, and (c) 0 m /s2.
You first have to find the acceleration of what is given. F1 has force of +8.0N, and box has mass of 4.7kg.
F = ma, so 8.0/4.7 is 1.7m/s2.
Now to view the other question, you'd use the same equation only this time, you need to use the newly acquired acceleration from above.
F2 = (4.7)(6.7-1.7) = 23.5N
F2 = (4.7)(-6.7-1.7) = -39.48N
F2 = (4.7)(0-1.7) = -7.99N
Well, well, well, let's try to solve this problem with a touch of humor, shall we?
First things first, let's calculate the net force acting on the box for each scenario. We have F1-> = +8.0 N, and we'll represent F2-> as +F2-> or -F2-> for left or right directions, respectively.
(a) When the acceleration is +6.7 m/s2, the net force acting on the box is given by the equation Fnet = ma. Plugging in the values, we get:
Fnet = (m)(a)
Fnet = (4.7 kg)(6.7 m/s2)
Fnet ≈ 31.49 N
Since the box is accelerating to the right, in order to balance out the forces, F2-> must be directed to the left. So, the magnitude of F2-> is equal to Fnet, approximately 31.49 N, and the direction is to the left.
(b) When the acceleration is -6.7 m/s2, we follow the same steps as before:
Fnet = (m)(a)
Fnet = (4.7 kg)(-6.7 m/s2)
Fnet ≈ -31.49 N
This time, the box is accelerating to the left, so F2-> must be directed to the right. Again, the magnitude of F2-> is approximately 31.49 N, and the direction is to the right.
(c) When the acceleration is 0 m/s2, the net force acting on the box is zero:
Fnet = (m)(a)
Fnet = (4.7 kg)(0 m/s2)
Fnet = 0 N
Since there is no acceleration, it means that F1-> is perfectly balanced by F2->. In this case, the magnitude of F2-> is also zero, and the direction doesn't matter. The box will remain at rest.
So, there you have it! The magnitude and direction of F2-> depend on the acceleration of the box. Keep in mind that this is a simplified scenario without friction. Thank you for the question, and remember to keep your forces balanced, just like a clown on a unicycle!
To find the magnitude and direction of F2->, we will use Newton's second law of motion, which states that the net force on an object equals its mass multiplied by its acceleration.
(a) Given:
F1-> = +8.0 N
m = 4.7 kg
a = +6.7 m/s^2 (since the acceleration is positive)
Using Newton's second law:
∑F = m * a
Since there is no friction, the net force is the sum of the individual forces acting on the box. In this case, the net force is F1-> + F2->.
So, we have:
F1-> + F2-> = m * a
Substituting the given values:
8.0 N + F2-> = 4.7 kg * (6.7 m/s^2)
Simplifying the equation:
F2-> = (4.7 kg * 6.7 m/s^2) - 8.0 N
Calculating:
F2-> = 31.49 N - 8.0 N
F2-> = 23.49 N
Therefore, the magnitude of F2-> when the acceleration is +6.7 m/s^2 is 23.49 N, and its direction depends on whether F2-> is pointing to the right or left.
(b) Given:
a = -6.7 m/s^2 (since the acceleration is negative)
Using the same equation:
F1-> + F2-> = m * a
Substituting the values:
8.0 N + F2-> = 4.7 kg * (-6.7 m/s^2)
Solving for F2->:
F2-> = (4.7 kg * -6.7 m/s^2) - 8.0 N
Calculating:
F2-> = -31.49 N - 8.0 N
F2-> = -39.49 N
Therefore, the magnitude of F2-> when the acceleration is -6.7 m/s^2 is 39.49 N, and its direction depends on whether F2-> is pointing to the right or left.
(c) Given:
a = 0 m/s^2 (since the acceleration is zero)
Using the same equation:
F1-> + F2-> = m * a
Substituting the values:
8.0 N + F2-> = 0 kg * 0 m/s^2
Since the acceleration is zero, the net force on the box must also be zero.
Simplifying the equation:
F2-> = 0 N - 8.0 N
F2-> = -8.0 N
Therefore, the magnitude of F2-> when the acceleration is 0 m/s^2 is 8.0 N, and its direction depends on whether F2-> is pointing to the right or left.
To find the magnitude and direction of F2->, 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.
(a) When the acceleration of the box is +6.7 m/s^2:
Since there is no friction, the net force is equal to the sum of the individual forces acting on the box.
F_net = F1 + F2
F2 = F_net - F1
The mass of the box, m = 4.7 kg
The acceleration, a = +6.7 m/s^2
The force F1 = +8.0 N
Substituting the given values into the equation:
F2 = (m*a) - F1
F2 = (4.7 kg * 6.7 m/s^2) - 8.0 N
(b) When the acceleration of the box is -6.7 m/s^2:
In this case, the net force is in the opposite direction to F1. Therefore, the equation becomes:
F2 = F_net - F1
F2 = (m*a) - F1
F2 = (4.7 kg * -6.7 m/s^2) - 8.0 N
(c) When the acceleration of the box is 0 m/s^2:
If the box is not accelerating (acceleration = 0), then the net force acting on the box is also 0. Therefore, the equation becomes:
F2 = F_net - F1
F2 = 0 N - 8.0 N
By substituting the given values into the equations, we can calculate the magnitude and direction of F2 for each scenario.