An object with a mass of #4 kg# is acted on by two forces. The first is #F_1= < -1 N , 3 N># and the second is #F_2 = < 6 N, 5 N>#. What is the object's rate and direction of acceleration?

The resultant force is

We apply Newton's second Law

The magnitude of the acceleration is

The angle lies in the 1st quadrant

The object's rate and direction of acceleration can be determined by first finding the net force acting on the object by summing the individual forces, and then using Newton's second law (F = ma) to calculate the acceleration vector.

Given:

• $m = 4 \, \text{kg}$
• $\vec{F}_1 = \langle -1 \, \text{N}, 3 \, \text{N} \rangle$
• $\vec{F}_2 = \langle 6 \, \text{N}, 5 \, \text{N} \rangle$

To find the net force: $\vec{F}_{\text{net}} = \vec{F}_1 + \vec{F}_2$ $\vec{F}_{\text{net}} = \langle -1 \, \text{N} + 6 \, \text{N}, \, 3 \, \text{N} + 5 \, \text{N} \rangle$ $\vec{F}_{\text{net}} = \langle 5 \, \text{N}, \, 8 \, \text{N} \rangle$

Now, using Newton's second law: $\vec{F}_{\text{net}} = m \cdot \vec{a}$ $\vec{a} = \frac{\vec{F}_{\text{net}}}{m}$ $\vec{a} = \frac{\langle 5 \, \text{N}, \, 8 \, \text{N} \rangle}{4 \, \text{kg}}$ $\vec{a} = \langle \frac{5}{4} \, \text{m/s}^2, \, \frac{8}{4} \, \text{m/s}^2 \rangle$ $\vec{a} = \langle 1.25 \, \text{m/s}^2, \, 2 \, \text{m/s}^2 \rangle$

So, the object's rate of acceleration is $1.25 \, \text{m/s}^2$ in the x-direction and $2 \, \text{m/s}^2$ in the y-direction.