Question

1. kg measures
2. n measures
3. m/s² measures
4. 1n=
5. the acceleration due to gravity on earth is and on the moon it is
6. what are the three formulas which describe the relationship between mass, force and acceleration?
7. a force of 52 n acts upon a 4 kg block sitting on the ground. calculate the acceleration of the object.
8. a 5 kg block is pulled across a table by a force of 61 n. calculate the acceleration of the object.
9. a roller coaster pushes a 25 kg person upward with a force of 300 n. what is the acceleration?
10. an object of mass 10 kg is accelerated upward at 2 m/s². what force is required?
11. what is the mass of an object if a force of 17 n causes it to accelerate at 1.5 m/s/s?
12. what is the acceleration of a 10 kg object if a force of 3 n is applied to it?
13. what is the mass of an object that requires a force of 25 n to accelerate at 5 m/s/s?
14. how much weight is generated by an 80 kg person on planet earth? on the moon?

Explanation:

Step1: Use Newton's second law

Newton's second law is \(F = ma\), where \(F\) is force, \(m\) is mass, and \(a\) is acceleration. We can rearrange this formula to find \(a=\frac{F}{m}\), \(m = \frac{F}{a}\)

Step2: Solve problem 22

Given \(F = 52N\) and \(m=4kg\). Using \(a=\frac{F}{m}\), we substitute the values: \(a=\frac{52}{4}\)

$$a = 13m/s^{2}$$

Step3: Solve problem 23

Given \(F = 61N\) and \(m = 5kg\). Using \(a=\frac{F}{m}\), we substitute the values: \(a=\frac{61}{5}\)

$$a=12.2m/s^{2}$$

Step4: Solve problem 24

Given \(F = 300N\) and \(m = 25kg\). Using \(a=\frac{F}{m}\), we substitute the values: \(a=\frac{300}{25}\)

$$a = 12m/s^{2}$$

Step5: Solve problem 25

Given \(m = 10kg\) and \(a=2m/s^{2}\). Using \(F=ma\), we substitute the values: \(F=10\times2\)

$$F = 20N$$

Step6: Solve problem 26

Given \(F = 17N\) and \(a = 1.5m/s^{2}\). Using \(m=\frac{F}{a}\), we substitute the values: \(m=\frac{17}{1.5}\)

$$m=\frac{34}{3}\approx11.33kg$$

Step7: Solve problem 27

Given \(F = 3N\) and \(m = 10kg\). Using \(a=\frac{F}{m}\), we substitute the values: \(a=\frac{3}{10}\)

$$a = 0.3m/s^{2}$$

Step8: Solve problem 28

Given \(F = 25N\) and \(a = 5m/s^{2}\). Using \(m=\frac{F}{a}\), we substitute the values: \(m=\frac{25}{5}\)

$$m = 5kg$$

Step9: Solve problem 29

On Earth, \(g = 9.8m/s^{2}\), \(m = 80kg\). Using \(F=mg\), \(F=80\times9.8\)

$$F = 784N$$

On the Moon, \(g=\frac{9.8}{6}\approx1.63m/s^{2}\), \(m = 80kg\). Using \(F=mg\), \(F=80\times1.63\)

$$F\approx130.4N$$

Answer:

1. \(13m/s^{2}\)
2. \(12.2m/s^{2}\)
3. \(12m/s^{2}\)
4. \(20N\)
5. \(\frac{34}{3}\approx11.33kg\)
6. \(0.3m/s^{2}\)
7. \(5kg\)
8. On Earth: \(784N\), On the Moon: \(130.4N\)