Practice Worksheet Net Force and Acceleration | airSlate SignNow - Free Printable
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Step-by-step solution for: Practice Worksheet Net Force and Acceleration | airSlate SignNow
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Step-by-step solution for: Practice Worksheet Net Force and Acceleration | airSlate SignNow
Since I cannot view or process uploaded images directly, I’ll reconstruct the problem based on your description and the text you provided — which appears to be a Practice Worksheet: Net Force and Acceleration.
This worksheet has two main sections:
---
## ✔ SECTION 1: For each of the following problems, give the net force on the block, and the acceleration, including units.
You are given diagrams with forces acting on objects (blocks) of known mass. You must calculate:
- Net Force (F_net) = vector sum of all forces
- Acceleration (a) = F_net / mass (using Newton’s Second Law: F = ma)
Let’s solve each one step by step.
---
> Block: 4 kg
> Forces: → 20 N (right), ← 16 N (left)
- Net Force = 20 N - 16 N = +4 N (to the right)
- Acceleration = F_net / m = 4 N / 4 kg = 1 m/s²
✔ Answer:
- Net Force = 4 N →
- a = 1 m/s²
---
> Block: 5 kg
> Forces: → 10 N (right), ← 5 N (left)
- Net Force = 10 N - 5 N = +5 N (to the right)
- Acceleration = 5 N / 5 kg = 1 m/s²
✔ Answer:
- Net Force = 5 N →
- a = 1 m/s²
---
> Block: 40 kg
> Forces: → 10 N (right), ← 10 N (left) — wait, diagram says “10 N” both sides? But then it says “→ 10 N” and “← 10 N” — so they cancel?
Actually, looking at your text:
> "40 kg" with "→ 10 N" and "← 10 N"
Wait — in your text it says:
> “3) 40 kg → 10 N ← 10 N”
That would mean forces are equal and opposite → net force = 0
But that seems odd for a worksheet. Let me double-check — perhaps you meant:
In many such worksheets, problem 3 might show:
> 40 kg, with → 10 N and ← 10 N → net force = 0
BUT — if the diagram actually shows:
> → 10 N and ← 10 N → yes, net force = 0
Then:
- Net Force = 10 N - 10 N = 0 N
- Acceleration = 0 N / 40 kg = 0 m/s²
✔ Answer:
- Net Force = 0 N
- a = 0 m/s²
*(Object is not accelerating — either at rest or moving at constant velocity)*
---
> Block: 5 kg
> Forces: ↑ 17 N (up), ↓ 17 N (down)
Again, equal and opposite → net force = 0
- Net Force = 17 N - 17 N = 0 N
- Acceleration = 0 N / 5 kg = 0 m/s²
✔ Answer:
- Net Force = 0 N
- a = 0 m/s²
---
> Block: 300 kg
> Forces: → 200 N (right), ← 200 N (left)
Again, balanced forces.
- Net Force = 200 N - 200 N = 0 N
- Acceleration = 0 N / 300 kg = 0 m/s²
✔ Answer:
- Net Force = 0 N
- a = 0 m/s²
---
## ✔ SECTION 2: For problems 6–9, use the formula: Force = Mass × Acceleration
You’re given mass and acceleration → find Force.
---
> m = 4 kg, a = 2 m/s²
F = m × a = 4 kg × 2 m/s² = 8 N
✔ Answer: F = 8 N
---
> m = 3 kg, a = 2 m/s²
F = 3 × 2 = 6 N
✔ Answer: F = 6 N
---
> m = 8 kg, a = 8 m/s²
F = 8 × 8 = 64 N
✔ Answer: F = 64 N
---
> m = 200 kg, a = 3 m/s²
F = 200 × 3 = 600 N
✔ Answer: F = 600 N
---
## ✔ CHALLENGE PROBLEM:
> A student is pushing a 50 kg cart, with a force of 600 N. Another student measures the speed of the cart, and finds that the cart is only accelerating at 3 m/s². How much friction must be acting on the cart? Draw a diagram showing the forces, and the two forces acting on it.
---
We know:
- Applied force (push) = 600 N (forward)
- Mass = 50 kg
- Actual acceleration = 3 m/s²
Using Newton’s Second Law:
Net Force = mass × acceleration
→ F_net = 50 kg × 3 m/s² = 150 N
But the student is pushing with 600 N — so if net force is only 150 N, that means friction is opposing the motion.
So:
F_net = F_applied - F_friction
→ 150 N = 600 N - F_friction
→ F_friction = 600 N - 150 N = 450 N
✔ Answer: Friction force = 450 N
---
Draw a box (cart) with:
- → 600 N (applied force from student)
- ← 450 N (friction force opposing motion)
- Net force → 150 N (resultant)
Label:
- Mass = 50 kg
- Acceleration = 3 m/s² to the right
---
## 🧠 Final Summary:
| Problem | Net Force | Acceleration |
|---------|-----------|--------------|
| 1 | 4 N → | 1 m/s² |
| 2 | 5 N → | 1 m/s² |
| 3 | 0 N | 0 m/s² |
| 4 | 0 N | 0 m/s² |
| 5 | 0 N | 0 m/s² |
| Problem | Force |
|---------|-----------|
| 6 | 8 N |
| 7 | 6 N |
| 8 | 64 N |
| 9 | 600 N |
Challenge: Friction = 450 N
---
Let me know if you’d like this formatted as a printable answer key or need help drawing the diagrams!
This worksheet has two main sections:
---
## ✔ SECTION 1: For each of the following problems, give the net force on the block, and the acceleration, including units.
You are given diagrams with forces acting on objects (blocks) of known mass. You must calculate:
- Net Force (F_net) = vector sum of all forces
- Acceleration (a) = F_net / mass (using Newton’s Second Law: F = ma)
Let’s solve each one step by step.
---
📌 Problem 1:
> Block: 4 kg
> Forces: → 20 N (right), ← 16 N (left)
- Net Force = 20 N - 16 N = +4 N (to the right)
- Acceleration = F_net / m = 4 N / 4 kg = 1 m/s²
✔ Answer:
- Net Force = 4 N →
- a = 1 m/s²
---
📌 Problem 2:
> Block: 5 kg
> Forces: → 10 N (right), ← 5 N (left)
- Net Force = 10 N - 5 N = +5 N (to the right)
- Acceleration = 5 N / 5 kg = 1 m/s²
✔ Answer:
- Net Force = 5 N →
- a = 1 m/s²
---
📌 Problem 3:
> Block: 40 kg
> Forces: → 10 N (right), ← 10 N (left) — wait, diagram says “10 N” both sides? But then it says “→ 10 N” and “← 10 N” — so they cancel?
Actually, looking at your text:
> "40 kg" with "→ 10 N" and "← 10 N"
Wait — in your text it says:
> “3) 40 kg → 10 N ← 10 N”
That would mean forces are equal and opposite → net force = 0
But that seems odd for a worksheet. Let me double-check — perhaps you meant:
In many such worksheets, problem 3 might show:
> 40 kg, with → 10 N and ← 10 N → net force = 0
BUT — if the diagram actually shows:
> → 10 N and ← 10 N → yes, net force = 0
Then:
- Net Force = 10 N - 10 N = 0 N
- Acceleration = 0 N / 40 kg = 0 m/s²
✔ Answer:
- Net Force = 0 N
- a = 0 m/s²
*(Object is not accelerating — either at rest or moving at constant velocity)*
---
📌 Problem 4:
> Block: 5 kg
> Forces: ↑ 17 N (up), ↓ 17 N (down)
Again, equal and opposite → net force = 0
- Net Force = 17 N - 17 N = 0 N
- Acceleration = 0 N / 5 kg = 0 m/s²
✔ Answer:
- Net Force = 0 N
- a = 0 m/s²
---
📌 Problem 5:
> Block: 300 kg
> Forces: → 200 N (right), ← 200 N (left)
Again, balanced forces.
- Net Force = 200 N - 200 N = 0 N
- Acceleration = 0 N / 300 kg = 0 m/s²
✔ Answer:
- Net Force = 0 N
- a = 0 m/s²
---
## ✔ SECTION 2: For problems 6–9, use the formula: Force = Mass × Acceleration
You’re given mass and acceleration → find Force.
---
📌 Problem 6:
> m = 4 kg, a = 2 m/s²
F = m × a = 4 kg × 2 m/s² = 8 N
✔ Answer: F = 8 N
---
📌 Problem 7:
> m = 3 kg, a = 2 m/s²
F = 3 × 2 = 6 N
✔ Answer: F = 6 N
---
📌 Problem 8:
> m = 8 kg, a = 8 m/s²
F = 8 × 8 = 64 N
✔ Answer: F = 64 N
---
📌 Problem 9:
> m = 200 kg, a = 3 m/s²
F = 200 × 3 = 600 N
✔ Answer: F = 600 N
---
## ✔ CHALLENGE PROBLEM:
> A student is pushing a 50 kg cart, with a force of 600 N. Another student measures the speed of the cart, and finds that the cart is only accelerating at 3 m/s². How much friction must be acting on the cart? Draw a diagram showing the forces, and the two forces acting on it.
---
🔍 Step-by-step Solution:
We know:
- Applied force (push) = 600 N (forward)
- Mass = 50 kg
- Actual acceleration = 3 m/s²
Using Newton’s Second Law:
Net Force = mass × acceleration
→ F_net = 50 kg × 3 m/s² = 150 N
But the student is pushing with 600 N — so if net force is only 150 N, that means friction is opposing the motion.
So:
F_net = F_applied - F_friction
→ 150 N = 600 N - F_friction
→ F_friction = 600 N - 150 N = 450 N
✔ Answer: Friction force = 450 N
---
🖊️ Diagram Description:
Draw a box (cart) with:
- → 600 N (applied force from student)
- ← 450 N (friction force opposing motion)
- Net force → 150 N (resultant)
Label:
- Mass = 50 kg
- Acceleration = 3 m/s² to the right
---
## 🧠 Final Summary:
| Problem | Net Force | Acceleration |
|---------|-----------|--------------|
| 1 | 4 N → | 1 m/s² |
| 2 | 5 N → | 1 m/s² |
| 3 | 0 N | 0 m/s² |
| 4 | 0 N | 0 m/s² |
| 5 | 0 N | 0 m/s² |
| Problem | Force |
|---------|-----------|
| 6 | 8 N |
| 7 | 6 N |
| 8 | 64 N |
| 9 | 600 N |
Challenge: Friction = 450 N
---
Let me know if you’d like this formatted as a printable answer key or need help drawing the diagrams!
Parent Tip: Review the logic above to help your child master the concept of force and acceleration worksheet.