✏️ Practice Force Problems

Apply Newton's Laws to real situations

📍 Overview 🔍 Observe 💡 Understand ✏️ Practice

📝 Force Problems

📐 Quick Reference:
F = ma (Newton's Second Law)
W = mg (Weight)
g = 10 m/s² (use for simplicity)
For inclined plane: mg sin θ (parallel), mg cos θ (perpendicular)
Problem 1.1 Beginner
A 5 kg box is pushed with a force of 20 N. What is its acceleration? (Ignore friction)

Step-by-Step Solution:

  • Given: m = 5 kg, F = 20 N
  • Formula: F = ma
  • Calculation: 20 = 5 × a
  • a = 20/5 = 4 m/s²
✅ Answer: 4 m/s²
🔍 Observation Connection: This is like pushing an empty box - it accelerates easily!
Problem 1.2 Beginner
What force is needed to accelerate a 2 kg object at 3 m/s²?

Step-by-Step Solution:

  • Given: m = 2 kg, a = 3 m/s²
  • Formula: F = ma
  • Calculation: F = 2 × 3 = 6 N
✅ Answer: 6 N
Problem 1.3 Beginner
A 10 kg object is falling freely near Earth's surface. What is the force of gravity on it? (g = 10 m/s²)

Step-by-Step Solution:

  • Given: m = 10 kg, g = 10 m/s²
  • Formula: Weight W = mg
  • Calculation: W = 10 × 10 = 100 N
✅ Answer: 100 N downward
🧠 Think Deeper: This is the force Earth exerts on the object. By Newton's Third Law, the object exerts an equal force upward on Earth!
Problem 2.1 Intermediate
A 15 kg box is pulled horizontally with a force of 50 N. If friction opposes motion with 10 N, find the acceleration.

Step-by-Step Solution:

  • Given: m = 15 kg, F_pull = 50 N, F_friction = 10 N
  • Net force: F_net = F_pull - F_friction = 50 - 10 = 40 N
  • Formula: F_net = ma
  • Calculation: 40 = 15 × a
  • a = 40/15 = 2.67 m/s²
✅ Answer: 2.67 m/s²
🔍 Observation Connection: This explains why it's harder to push a box on rough ground - friction opposes motion!
Problem 2.2 Intermediate
A 2 kg book rests on a table. What is the normal force exerted by the table? (g = 10 m/s²)

Step-by-Step Solution:

  • Given: m = 2 kg, g = 10 m/s²
  • Forces on book: Weight down = mg = 20 N, Normal force up = N
  • Since book is at rest: Net force = 0
  • N - mg = 0
  • N = mg = 20 N
✅ Answer: 20 N upward
🔍 Observation Connection: This is exactly Observation 4! The table pushes up with exactly the same force gravity pulls down.
Problem 2.3 Intermediate
A 60 kg person stands in an elevator. Find the normal force when: (a) elevator at rest, (b) elevator accelerating upward at 2 m/s², (c) elevator accelerating downward at 2 m/s². (g = 10 m/s²)

Step-by-Step Solution:

  • Given: m = 60 kg, g = 10 m/s², weight = mg = 600 N
  • Formula: F_net = ma = N - mg (taking up as positive)
  • Case (a) - At rest: a = 0, so N - mg = 0, N = mg = 600 N
  • Case (b) - Upward acceleration (+2 m/s²): N - 600 = 60 × 2
  • N - 600 = 120
  • N = 720 N (you feel heavier)
  • Case (c) - Downward acceleration (-2 m/s²): N - 600 = 60 × (-2)
  • N - 600 = -120
  • N = 480 N (you feel lighter)
✅ Answer: (a) 600 N, (b) 720 N, (c) 480 N
🔍 Observation Connection: This explains why you feel heavier when elevator starts going up, and lighter when it goes down!
Problem 3.1 Advanced
Two masses (m₁ = 4 kg, m₂ = 6 kg) are connected by a string over a frictionless pulley. Find the acceleration of the system and tension in the string. (g = 10 m/s²)

📐 Atwood Machine Diagram

        ○───────○
        │       │
        │       │
        │       │
        │       │
       [ ]     [ ]
       m₁=4    m₂=6
        ↓       ↓
      a↑?      a↓?
m₁= 4 kg (lighter)
m₂= 6 kg (heavier)
g= 10 m/s²
Directionm₁ ↑, m₂ ↓
🔧 How it works: The heavier mass (m₂) accelerates downward, pulling the lighter mass (m₁) upward. Tension is same throughout string because pulley is frictionless.

Step-by-Step Solution:

  • For m₁ (lighter, accelerates UP): T - m₁g = m₁a
  • T - 40 = 4a ...(1)
  • For m₂ (heavier, accelerates DOWN): m₂g - T = m₂a
  • 60 - T = 6a ...(2)
  • Add equations (1) and (2): (T - 40) + (60 - T) = 4a + 6a
  • 20 = 10a
  • a = 2 m/s²
  • Find T using equation (1): T - 40 = 4 × 2
  • T - 40 = 8
  • T = 48 N
✅ Answer: a = 2 m/s² (m₁↑, m₂↓), T = 48 N
🧠 Check: Does this make sense? For m₂: 60 - 48 = 12 N net force, 12/6 = 2 m/s² ✓
Problem 3.2 Advanced
A 5 kg block is on a frictionless 30° incline. Find: (a) acceleration down the incline, (b) normal force.

📐 Inclined Plane Diagram

            ╱│
           ╱ │
          ╱  │
         ╱   │ [5kg]
        ╱    │
       ╱30°  │
      ╱──────┘
      
      Force components:
      ↓ mg = 50 N
      ╱  ╲
     ╱    ╲
    ╱      ╲
   ╱        ╲
  ╱ 30°      ╲ mg sin θ
 ─────────────────
 mg cos θ
m= 5 kg
θ= 30°
g= 10 m/s²
mg= 50 N
🔧 Force Components:
• Parallel to incline: mg sin θ = 50 × 0.5 = 25 N (down incline)
• Perpendicular to incline: mg cos θ = 50 × 0.866 = 43.3 N (into incline)

Step-by-Step Solution:

  • Given: m = 5 kg, θ = 30°, g = 10 m/s²
  • Weight components:
    • Parallel to incline: mg sin θ = 5 × 10 × sin 30°
    • = 50 × 0.5 = 25 N
    • Perpendicular to incline: mg cos θ = 5 × 10 × cos 30°
    • = 50 × 0.866 = 43.3 N
  • Part (a) - Acceleration down incline:
    • F_parallel = ma
    • 25 = 5a
    • a = 5 m/s²
  • Part (b) - Normal force:
    • N = mg cos θ (perpendicular component)
    • N = 43.3 N
✅ Answer: (a) a = 5 m/s² down incline, (b) N = 43.3 N
🧠 Think Deeper: Notice acceleration (5 m/s²) is less than g (10 m/s²)! Only part of gravity causes acceleration down incline.
🧠 Critical Thinking Challenge:

In Problem 3.1 (Atwood machine), what would happen if m₁ = m₂? What would be the acceleration and tension? Can you prove it mathematically?

Hint: If masses are equal, system is balanced. What does F = ma tell you?

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