Two horizontal forces F1 and F2, are pulling an object of mass 3.05kg from two opposite sides. F1 pulls to the right and F2 pulls to the left. The magnitude of F1 is 99.2N. The object moves strictly along the horizontal x-axis, which we chose as positive to the right. Find the magnitude of F2 if the object's horizontal acceleration is 7.21m/s^2.
b). Find the magnitude of F2 if the acceleration is 0m/s^2.
c).Find the magnitude of F2 if the acceleration is -7.21m/s^2.
(a) ma=F1-F2
F2=F1-ma
(b)F2=F1
(c) -ma=F1-F2
F2=F1+ma
Matatatamag
To find the magnitude of F2 in each scenario, 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. In equation form, it is represented as:
F_net = m * a
In this case, we know the mass of the object is 3.05 kg and the acceleration in each scenario. Let's go through each scenario one by one.
a) When the acceleration is 7.21 m/s^2:
We know that F_net is the difference between F1 and F2, as the forces are in opposite directions. So, we have:
F_net = F1 - F2
Substituting the given values, we have:
F1 = 99.2 N
m = 3.05 kg
a = 7.21 m/s^2
From Newton's second law, we can rearrange the equation to solve for F2:
F2 = F1 - F_net
F2 = F1 - m * a
Now substituting the given values, we get:
F2 = 99.2 N - (3.05 kg * 7.21 m/s^2)
Calculating this, we find:
F2 ≈ 99.2 N - 22.0855 N
F2 ≈ 77.1145 N
Therefore, the magnitude of F2 when the acceleration is 7.21 m/s^2 is approximately 77.1145 N.
b) When the acceleration is 0 m/s^2:
In this case, the net force acting on the object is zero, as there is no acceleration. So we have:
F_net = 0
Again, using Newton's second law, we can find F2:
F2 = F1 - F_net
F2 = F1 - 0
F2 = F1
Therefore, the magnitude of F2 when the acceleration is 0 m/s^2 is equal to F1, which is given to be 99.2 N.
c) When the acceleration is -7.21 m/s^2:
Since the acceleration is negative, it means the object is experiencing deceleration or moving in the opposite direction. In this case, the net force would be in the direction of F2. So:
F_net = F2 - F1
Substituting the given values, we get:
F_net = F2 - F1
F_net = m * a
Rearranging the equation to solve for F2, we have:
F2 = F_net + F1
F2 = m * a + F1
Substituting the given values, we calculate:
F2 = (3.05 kg * -7.21 m/s^2) + 99.2 N
Simplifying this equation gives:
F2 ≈ -22.0855 N + 99.2 N
F2 ≈ 77.1145 N
Therefore, the magnitude of F2 when the acceleration is -7.21 m/s^2 is approximately 77.1145 N.