Question

1. how much force is required to accelerate a 2kg mass at 3 m/s2?
2. what is the mass of an object that requires 100n of force in order to accelerate it at 10m/s²?
3. what is the acceleration of a 10kg mass pushed by a 5n force?
4. given a force of 88n and an acceleration of 4m/s2, what is the mass?
5. how much force is required to accelerate a 12kg mass at 5m/s2?
6. given a force of 10n and an acceleration of 5m/s2. what is the mass?
7. how much force is required to accelerate a 5kg mass at 20m/s2?
8. what is the acceleration of a 5kg mass pushed by a 10n force?
9. how much force is required to accelerate an 8kg mass at 5m/s2?
10. find the acceleration of the 2kg block in the following diagram.

4n 8n

Explanation:

Step1: Recall Newton's second - law formula

$F = ma$, where $F$ is force in Newtons (N), $m$ is mass in kilograms (kg), and $a$ is acceleration in meters per second squared ($m/s^{2}$). We can also re - arrange it as $m=\frac{F}{a}$ and $a = \frac{F}{m}$.

Step2: Solve problem 1

Given $m = 2kg$ and $a=3m/s^{2}$, using $F = ma$, we have $F=2\times3 = 6N$.

Step3: Solve problem 2

Given $F = 100N$ and $a = 10m/s^{2}$, using $m=\frac{F}{a}$, we get $m=\frac{100}{10}=10kg$.

Step4: Solve problem 3

Given $m = 10kg$ and $F = 5N$, using $a=\frac{F}{m}$, we obtain $a=\frac{5}{10}=0.5m/s^{2}$.

Step5: Solve problem 4

Given $F = 88N$ and $a = 4m/s^{2}$, using $m=\frac{F}{a}$, we find $m=\frac{88}{4}=22kg$.

Step6: Solve problem 5

Given $m = 12kg$ and $a = 5m/s^{2}$, using $F = ma$, we have $F=12\times5 = 60N$.

Step7: Solve problem 6

Given $F = 10N$ and $a = 5m/s^{2}$, using $m=\frac{F}{a}$, we get $m=\frac{10}{5}=2kg$.

Step8: Solve problem 7

Given $m = 5kg$ and $a = 20m/s^{2}$, using $F = ma$, we obtain $F=5\times20 = 100N$.

Step9: Solve problem 8

Given $m = 5kg$ and $F = 10N$, using $a=\frac{F}{m}$, we find $a=\frac{10}{5}=2m/s^{2}$.

Step10: Solve problem 9

Given $m = 8kg$ and $a = 5m/s^{2}$, using $F = ma$, we have $F=8\times5 = 40N$.

Step11: Solve problem 10

First, find the net force. The forces acting on the block are $F_{1}=4N$ and $F_{2}=8N$. The net force $F_{net}=8 - 4=4N$ (assuming the 8N force is in the positive direction). Given $m = 2kg$, using $a=\frac{F_{net}}{m}$, we get $a=\frac{4}{2}=2m/s^{2}$.

Answer:

1. 6N
2. 10kg
3. $0.5m/s^{2}$
4. 22kg
5. 60N
6. 2kg
7. 100N
8. $2m/s^{2}$
9. 40N
10. $2m/s^{2}$