A net force of [tex][tex]$6.8 \, N$[/tex][/tex] accelerates a [tex][tex]$31 \, kg$[/tex][/tex] scooter across a level parking lot. What is the magnitude of the scooter's acceleration? [tex](F = ma)[/tex]

A. [tex]0.22 \, m/s^2[/tex]
B. [tex]0.69 \, m/s^2[/tex]
C. [tex]3.2 \, m/s^2[/tex]
D. [tex]4.6 \, m/s^2[/tex]



Answer :

To solve this problem, we need to use Newton's Second Law of Motion, which is expressed by the formula:

[tex]\[ F = ma \][/tex]

where:
- [tex]\( F \)[/tex] is the net force applied to the object, measured in Newtons (N).
- [tex]\( m \)[/tex] is the mass of the object, measured in kilograms (kg).
- [tex]\( a \)[/tex] is the acceleration of the object, measured in meters per second squared (m/s²).

We are given:
- A net force [tex]\( F = 6.8 \)[/tex] N
- A mass [tex]\( m = 31 \)[/tex] kg

We are required to find the magnitude of the scooter's acceleration [tex]\( a \)[/tex].

Step-by-step solution:

1. Identify the formula to use:
Since we are given the force and mass, we use the formula [tex]\( F = ma \)[/tex].

2. Rearrange the formula to solve for acceleration ([tex]\( a \)[/tex]):
[tex]\[ a = \frac{F}{m} \][/tex]

3. Substitute the given values into the formula:
[tex]\[ a = \frac{6.8 \text{ N}}{31 \text{ kg}} \][/tex]

4. Perform the division to find the acceleration:
[tex]\[ a \approx 0.21935483870967742 \text{ m/s}^2 \][/tex]

5. Round the answer to two decimal places (as typical in physics for clarity):
[tex]\[ a \approx 0.22 \text{ m/s}^2 \][/tex]

Thus, the magnitude of the scooter's acceleration is approximately [tex]\( 0.22 \text{ m/s}^2 \)[/tex].

So, the correct answer is:
[tex]\[ \text{a} \quad 0.22 \text{ m/s}^2 \][/tex]