An object with a mass of #4 kg# is acted on by two forces. The first is #F_1= < 6 N , -2 N># and the second is #F_2 = < -5 N, 3 N>#. What is the object's rate and direction of acceleration?

as the two components of net force are equal,

To find the object's rate and direction of acceleration, we need to calculate the net force acting on the object and then apply Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.

Given the forces ( \mathbf{F_1} = \langle 6 , \text{N}, -2 , \text{N} \rangle ) and ( \mathbf{F_2} = \langle -5 , \text{N}, 3 , \text{N} \rangle ), we can find the net force by summing the individual force components:

Net force in the x-direction: ( F_{net_x} = F_{1x} + F_{2x} = 6 , \text{N} + (-5 , \text{N}) = 1 , \text{N} )

Net force in the y-direction: ( F_{net_y} = F_{1y} + F_{2y} = (-2 , \text{N}) + 3 , \text{N} = 1 , \text{N} )

Now, we have the net force components: ( \langle 1 , \text{N}, 1 , \text{N} \rangle ).

Next, we apply Newton's second law:

Acceleration in the x-direction: ( a_x = \frac{F_{net_x}}{m} = \frac{1 , \text{N}}{4 , \text{kg}} = 0.25 , \text{m/s}^2 )

Acceleration in the y-direction: ( a_y = \frac{F_{net_y}}{m} = \frac{1 , \text{N}}{4 , \text{kg}} = 0.25 , \text{m/s}^2 )

Therefore, the object's rate of acceleration is ( 0.25 , \text{m/s}^2 ) in both the x-direction and y-direction. As for the direction, it's important to note that acceleration is a vector quantity. The direction of acceleration can be described using angles or trigonometric functions, but in this case, the acceleration's direction aligns with the resultant force direction, which is ( 45^\circ ) relative to the positive x-axis.