The value 1.22 in this diagram seems to come from calculating the mass (m) of the object.

From the net force equation:

$F_{\text{net}} = m \cdot a$

We can rearrange to solve for mass $m$:

$m = \frac{F_{\text{net}}}{a}$

However, in the diagram, the calculation uses the mass directly to find the acceleration, and this value is given as 1.22. Let's verify if this mass is derived from the other forces in the problem.

1. The forces in the problem:

   • $F_{\text{app}} = 30 \, \text{N}$ (applied force)
   • $F_{\text{friction}} = 6 \, \text{N}$ (frictional force, calculated using $F_f = \mu \cdot F_N$)
   • $F_{\text{net}} = F_{\text{app}} - F_{\text{friction}} = 30 - 6 = 24 \, \text{N}$

2. Acceleration $a$: The acceleration is given as $4.67 \, \text{m/s}^2$, calculated from:

   $a = \frac{F_{\text{net}}}{m}$

3. Mass $m$: Rearranging the formula to solve for mass:

   $m = \frac{F_{\text{net}}}{a} = \frac{24}{4.67} \approx 1.22 \, \text{kg}$

Thus, the 1.22 comes from the calculated mass of the object based on the net force and the acceleration.

Do you have more questions or need clarification? Here are 5 related questions:

1. How is the frictional force $F_f$ calculated using $\mu \cdot F_N$?
2. What would change in the calculation if the coefficient of friction $\mu$ were different?
3. How is the net force $F_{\text{net}}$ related to the applied and frictional forces?
4. What is the significance of the acceleration value $4.67 \, \text{m/s}^2$?
5. How would the problem change if $F_{\text{app}}$ were reduced?

Tip: Always verify the net force equation by summing up all forces acting on the object along the direction of motion!