Question

part two: acceleration and force

1. calculate the force required to accelerate a baseball (0.142 kg) at each listed acceleration.

baseball mass (kg) acceleration (m/s²) force in newtons (n)
0.142 2
0.142 4
0.142 6
0.142 8
0.142 10
0.142 12
0.142 14
0.142 16
0.142 18
0.142 20
round to the nearest tenth.

Explanation:

Step1: Recall Newton's second - law formula

$F = ma$, where $F$ is force, $m$ is mass, and $a$ is acceleration.

Step2: Calculate force for each acceleration value

For $a = 2\ m/s^{2}$, $F=0.142\times2 = 0.284\ N\approx0.3\ N$
For $a = 4\ m/s^{2}$, $F=0.142\times4 = 0.568\ N\approx0.6\ N$
For $a = 6\ m/s^{2}$, $F=0.142\times6 = 0.852\ N\approx0.9\ N$
For $a = 8\ m/s^{2}$, $F=0.142\times8 = 1.136\ N\approx1.1\ N$
For $a = 10\ m/s^{2}$, $F=0.142\times10 = 1.42\ N\approx1.4\ N$
For $a = 12\ m/s^{2}$, $F=0.142\times12 = 1.704\ N\approx1.7\ N$
For $a = 14\ m/s^{2}$, $F=0.142\times14 = 1.988\ N\approx2.0\ N$
For $a = 16\ m/s^{2}$, $F=0.142\times16 = 2.272\ N\approx2.3\ N$
For $a = 18\ m/s^{2}$, $F=0.142\times18 = 2.556\ N\approx2.6\ N$
For $a = 20\ m/s^{2}$, $F=0.142\times20 = 2.84\ N\approx2.8\ N$

Answer:

Baseball Mass (kg)	Acceleration ($m/s^{2}$)	Force in Newtons (N)
0.142	4	0.6
0.142	6	0.9
0.142	8	1.1
0.142	10	1.4
0.142	12	1.7
0.142	14	2.0
0.142	16	2.3
0.142	18	2.6
0.142	20	2.8