Newton's Second Law of Motion worksheet with practice problems and solutions.

Let’s solve each problem step by step using Newton’s Second Law: F = m × a

We’ll go one question at a time.

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Problem 1:
A llama weighing 200 kg needs to be accelerated upward at 0.65 m/s². What force is needed?

→ We use F = m × a
→ m = 200 kg
→ a = 0.65 m/s²
→ F = 200 × 0.65 = 130 N

✔ Answer: 130 N

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Problem 2:
What net force is needed to accelerate a 1700 kg car forward at 3.5 m/s²?

→ F = m × a
→ m = 1700 kg
→ a = 3.5 m/s²
→ F = 1700 × 3.5

Let’s calculate:
1700 × 3 = 5100
1700 × 0.5 = 850
Total = 5100 + 850 = 5950 N

✔ Answer: 5950 N

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Problem 3:
A ball accelerates downward at 9.8 m/s² with a force of 12 N. What is its mass?

→ We rearrange the formula: m = F ÷ a
→ F = 12 N
→ a = 9.8 m/s²
→ m = 12 ÷ 9.8 ≈ 1.22 kg

But looking at the options:
- 0.8 kg
- 1.2 kg ← closest match
- 1.4 kg

Since 12 ÷ 9.8 is approximately 1.22, and 1.2 kg is the closest option, we choose 1.2 kg

✔ Answer: 1.2 kg

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Problem 4:
What is the acceleration of a 650 kg racing camel if the net force is 897 N?

→ Use a = F ÷ m
→ F = 897 N
→ m = 650 kg
→ a = 897 ÷ 650

Let’s divide:
897 ÷ 650 = ?

650 × 1.3 = 845
897 - 845 = 52
52 ÷ 650 = 0.08
So total ≈ 1.3 + 0.08 = 1.38 m/s²

Looking at options:
- 1.69 m/s²
- 1.38 m/s² ← matches
- 1.02 m/s²

✔ Answer: 1.38 m/s²

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Final Answer:
130 N, 5950 N, 1.2 kg, 1.38 m/s²