a 1.00-kg object is observed to accelerate at 10.0m/s∧2 in a direction 30.0 north of east. the force F2 acting on the object has magnitude 5.00 N and is directed north. determine the magnitude and direction of the force F1 acting on the object.

IF AN OBJECTS IS OBSEVED TO ACCELERATE

To determine the magnitude and direction of the force F1 acting on the object, we'll 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:

F_net = m * a

First, let's break down the forces acting on the object:

1. Force F1: The magnitude and direction of this force are what we need to determine.
2. Force F2: Magnitude = 5.00 N (given), Direction = North.

Given that the object is accelerating at 10.0 m/s^2, we know that the net force acting on it is:

F_net = m * a
F_net = (1.00 kg) * (10.0 m/s^2)
F_net = 10.0 N

Since F_net is the vector sum of F1 and F2, we can break it down into its components:

F_net_x = F1_x + F2_x
F_net_y = F1_y + F2_y

To simplify the problem, let's convert everything into a Cartesian coordinate system:

Positive x-axis: East
Positive y-axis: North

Since F2 is directed north, its components are:

F2_x = 0 (no component along the x-axis)
F2_y = +5.00 N (positive component along the y-axis)

Now, we need to find the components of the net force:

F_net_x = F1_x + F2_x
F_net_y = F1_y + F2_y

Since the object is accelerating 30.0° north of east, we can find the x and y components of the acceleration:

a_x = a * cos(30°)
a_y = a * sin(30°)

a_x = (10.0 m/s^2) * cos(30°)
a_y = (10.0 m/s^2) * sin(30°)

a_x = 8.66 m/s^2
a_y = 5.00 m/s^2

Now, let's calculate the x-component of the net force:

F_net_x = F1_x + F2_x
8.66 N = F1_x + 0
F1_x = 8.66 N (force component in the positive x direction)

Next, let's calculate the y-component of the net force:

F_net_y = F1_y + F2_y
5.00 N = F1_y + 5.00 N
F1_y = 0 N (force component in the positive y direction)

Finally, we can determine the magnitude and direction of F1 using the components we found:

F1 = √(F1_x^2 + F1_y^2)
F1 = √(8.66 N^2 + 0 N^2)
F1 = 8.66 N (magnitude of F1)

The direction of F1 can be calculated using the tangent function:

θ = tan^(-1)(F1_y / F1_x)
θ = tan^(-1)(0 N / 8.66 N)
θ = 0° (direction of F1)

Therefore, the magnitude of the force F1 is 8.66 N, and its direction is 0°.