Question: Watch the video and answer the 3 questions
Question 1: What equation represents Newton's 2nd Law relating force, mass, and acceleration?
Question 2: How much pulling force does a 100g mass pull the trolley with ?
Question 3: Fill in the missing average acceleration values
mass
(g)
Force
(N)
acceleration (m/s²)
1
2
3
average
100
1.0
1.263
1.253
1.257
1.258
80
0.8
0.994
1.002
0.996
60
0.6
0.736
0.728
0.737
0.734
40
0.4
0.473
0.462
0.459
20
0.2
0.227
0.219
0.208
0.218
F=ma - Science GCSE Physics Required Practical - Newton's 2nd Law
1. **State the problem:** We need to answer three questions related to Newton's 2nd Law and the given data table.
2. **Newton's 2nd Law formula:** The law states that the force $F$ acting on an object is equal to the mass $m$ of the object multiplied by its acceleration $a$.
$$F = ma$$
This means force, mass, and acceleration are directly related.
3. **Answer to Question 1:** The equation representing Newton's 2nd Law is:
$$F = ma$$
4. **Answer to Question 2:** Given a mass of $100\text{ g} = 0.1\text{ kg}$ and the force from the table is $1.0\text{ N}$.
So, the pulling force the $100\text{ g}$ mass exerts on the trolley is:
$$F = 1.0\text{ N}$$
5. **Answer to Question 3:** Fill in the missing average acceleration values by calculating the average of the three acceleration measurements for each mass.
- For $80\text{ g}$ mass:
$$\text{average} = \frac{0.994 + 1.002 + 0.996}{3} = \frac{2.992}{3} = 0.997$$
- For $40\text{ g}$ mass:
$$\text{average} = \frac{0.473 + 0.462 + 0.459}{3} = \frac{1.394}{3} = 0.465$$
6. **Summary of average accelerations:**
| Mass (g) | Average Acceleration (m/s²) |
|----------|-----------------------------|
| 100 | 1.258 |
| 80 | 0.997 |
| 60 | 0.734 |
| 40 | 0.465 |
| 20 | 0.218 |
These averages complete the missing values in the table.
**Final answers:**
1. Newton's 2nd Law equation: $$F = ma$$
2. Pulling force for $100\text{ g}$ mass: $$1.0\text{ N}$$
3. Missing average accelerations:
- For $80\text{ g}$: $$0.997\text{ m/s}^2$$
- For $40\text{ g}$: $$0.465\text{ m/s}^2$$