Physics worksheet featuring Newton's second law problems with force, mass, and acceleration calculations.

It looks like you've uploaded a physics worksheet involving Newton's Second Law of Motion, where the goal is to calculate net force and acceleration using the formula:

\[
F_{\text{net}} = m \cdot a
\]

Let’s go through the problems step by step based on the image you provided. I’ll explain how to solve each one.

---

Problem 1:

- Mass: \( m = 40 \, \text{kg} \)
- Acceleration: \( a = 2 \, \text{m/s}^2 \)

Solution:
\[
F_{\text{net}} = m \cdot a = 40 \, \text{kg} \times 2 \, \text{m/s}^2 = 80 \, \text{N}
\]

✔ Answer: Net Force = \( 80 \, \text{N} \)

---

Problem 2:

- Mass: \( m = 30 \, \text{kg} \)
- Net Force: \( F_{\text{net}} = 60 \, \text{N} \)

We need to find acceleration:
\[
a = \frac{F_{\text{net}}}{m} = \frac{60 \, \text{N}}{30 \, \text{kg}} = 2 \, \text{m/s}^2
\]

✔ Answer: Acceleration = \( 2 \, \text{m/s}^2 \)

---

Problem 3:

- Mass: \( m = 50 \, \text{kg} \)
- Net Force: \( F_{\text{net}} = 100 \, \text{N} \)

Find acceleration:
\[
a = \frac{100 \, \text{N}}{50 \, \text{kg}} = 2 \, \text{m/s}^2
\]

✔ Answer: Acceleration = \( 2 \, \text{m/s}^2 \)

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Problem 4:

- Mass: \( m = 75 \, \text{kg} \)
- Acceleration: \( a = 1.5 \, \text{m/s}^2 \)

Find net force:
\[
F_{\text{net}} = 75 \times 1.5 = 112.5 \, \text{N}
\]

✔ Answer: Net Force = \( 112.5 \, \text{N} \)

---

Problem 5:

- Mass: \( m = 120 \, \text{kg} \)
- Net Force: \( F_{\text{net}} = 120 \, \text{N} \)

Find acceleration:
\[
a = \frac{120}{120} = 1 \, \text{m/s}^2
\]

✔ Answer: Acceleration = \( 1 \, \text{m/s}^2 \)

---

Problem 6:

- Mass: \( m = 10 \, \text{kg} \)
- Net Force: \( F_{\text{net}} = 25 \, \text{N} \)

Find acceleration:
\[
a = \frac{25}{10} = 2.5 \, \text{m/s}^2
\]

✔ Answer: Acceleration = \( 2.5 \, \text{m/s}^2 \)

---

Problem 7:

- Mass: \( m = 50 \, \text{kg} \)
- Net Force: \( F_{\text{net}} = 200 \, \text{N} \)

Find acceleration:
\[
a = \frac{200}{50} = 4 \, \text{m/s}^2
\]

✔ Answer: Acceleration = \( 4 \, \text{m/s}^2 \)

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Problem 8:

- Mass: \( m = 12 \, \text{kg} \)
- Acceleration: \( a = 4 \, \text{m/s}^2 \)

Find net force:
\[
F_{\text{net}} = 12 \times 4 = 48 \, \text{N}
\]

✔ Answer: Net Force = \( 48 \, \text{N} \)

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Problem 9:

- Mass: \( m = 20 \, \text{kg} \)
- Net Force: \( F_{\text{net}} = 1200 \, \text{N} \)

Find acceleration:
\[
a = \frac{1200}{20} = 60 \, \text{m/s}^2
\]

✔ Answer: Acceleration = \( 60 \, \text{m/s}^2 \)

---

Challenge Problem (10):

> A student is pushing a 50 kg box with a force of 600 N. Another student measures the speed of the cart, and finds that the cart is only accelerating at 10 m/s². How much friction must be acting on the cart? Draw a free-body diagram and show your work.

#### Step 1: Use Newton’s Second Law
The net force is:
\[
F_{\text{net}} = m \cdot a = 50 \, \text{kg} \times 10 \, \text{m/s}^2 = 500 \, \text{N}
\]

But the applied force is 600 N. So the difference must be due to friction opposing motion.

\[
F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}
\]
\[
500 = 600 - F_{\text{friction}}
\]
\[
F_{\text{friction}} = 600 - 500 = 100 \, \text{N}
\]

✔ Answer: Friction = \( 100 \, \text{N} \)

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✔ Summary of Answers:

| Problem | Answer |
|--------|--------|
| 1 | 80 N |
| 2 | 2 m/s² |
| 3 | 2 m/s² |
| 4 | 112.5 N |
| 5 | 1 m/s² |
| 6 | 2.5 m/s² |
| 7 | 4 m/s² |
| 8 | 48 N |
| 9 | 60 m/s² |
| 10 | 100 N (friction) |

---

🔍 Key Concept:

Newton’s Second Law:
\[
F_{\text{net}} = m \cdot a
\]
Always remember:
- If two forces are acting (like applied and friction), use:
\[
F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}
\]
- Direction matters — assume forward as positive.

Let me know if you'd like a free-body diagram for the challenge problem!