To analyze this problem, let's determine the net force ($\Sigma F$) and the acceleration ($a$) of the object with a mass $m = 8 \, \text{kg}$.

Step 1: Breaking Down the Forces

The forces are given as:

1. Vertical Forces: $12.3 \, \text{N}$ upward and $12.3 \, \text{N}$ downward.
2. Horizontal Forces: $4.3 \, \text{N}$ to the left and $16 \, \text{N}$ to the left.

Since the upward and downward forces are equal, they cancel each other out. Therefore, there is no net force in the vertical direction.

Step 2: Calculating the Horizontal Net Force

In the horizontal direction:

• Total force to the left = $4.3 \, \text{N} + 16 \, \text{N} = 20.3 \, \text{N}$

Since there is no opposing horizontal force to the right, the net force $\Sigma F$ is: $\Sigma F = 20.3 \, \text{N} \quad \text{(to the left)}$

Step 3: Calculating Acceleration

Using Newton's second law, $F = ma$, we can solve for $a$: $a = \frac{\Sigma F}{m} = \frac{20.3 \, \text{N}}{8 \, \text{kg}} = 2.5375 \, \text{m/s}^2$

Rounding to two decimal places: $a \approx 2.54 \, \text{m/s}^2$

Summary

• Net Force (ΣF): $20.3 \, \text{N}$ to the left
• Acceleration (a): $2.54 \, \text{m/s}^2$ to the left

Would you like more details on any part of this solution?

Here are 5 related questions that build on this information:

1. What would happen to the acceleration if the mass increased to 10 kg?
2. How would the results change if there were an additional horizontal force to the right?
3. What is the significance of balanced forces in determining motion?
4. How would we solve for acceleration if there were forces at an angle?
5. What if friction were involved—how would that affect the net force?

Tip: When forces are balanced in one direction, the net force and acceleration will only depend on the forces in the unbalanced direction.