What is the work done when a constant force of 20 N is applied... | General Science MCQ | Examoogle

General SciencePhysics: Work, Energy and Power

What is the work done when a constant force of 20 N is applied at an angle of 30° to the direction of displacement of 13 m?

390 Joules

225 Joules

260 Joules

185 Joules

Correct Answer

🔬 SC • Science Concept & PrincipleGeneral SciencePhysics: Work, Energy and Power

Option B

225 Joules

Quick Summary:

The work done by a force is calculated as the dot product of the force vector and the displacement vector. For a constant force applied at an angle, the work is $W = F \cdot s \cdot \cos(\theta)$ , where $F = 20\text{ N}$ , $s = 13\text{ m}$ , and $\theta = 30°$ .

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🔬SCScience SolutionConcept & Principle

💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

Think of pulling a suitcase with wheels; pulling the handle at an angle means only part of your force actually moves the suitcase forward, while the rest just lifts it slightly.

🧠 Memory Mnemonic / Shortcut Aid

WFS Cos-Theta: Work is Force times displacement times the Cosine of the angle.

🔢 Key Formulas

$W = F \cdot s \cdot \cos(\theta)$ — Work done formula where $\theta$ is the angle between force and displacement vectors.

⚙️ Working Principle

Work is defined as the component of force acting in the direction of displacement multiplied by the displacement. Since $\cos(30°) = \frac{\sqrt{3}}{2} \approx 0.866$ , the calculation yields $20 \times 13 \times 0.866 \approx 225.16\text{ J}$ . This physical principle illustrates that only the force component parallel to the direction of motion contributes to work.

📌 Key Points

▸
Work is a scalar quantity.
▸
If the force is perpendicular to displacement (90 degrees), the work done is zero.
▸
The SI unit of work is Joule (J), equivalent to $1\text{ N}\cdot\text{m}$ .

✅ Advantages

▸
Vector analysis allows precise calculation of energy transfer in mechanical systems.

❌ Disadvantages / Limitations

▸
This formula assumes a constant force and a straight-line path.

🛠️ Applications / Uses

▸
Calculating power consumption in machinery.
▸
Analyzing energy conservation in physics problems.

📄 Additional Information

▸
Standard value: $\cos(30°) = \frac{\sqrt{3}}{2} \approx 0.866025$ .
▸
Calculation: $20 \times 13 \times 0.866025 = 225.166\text{ J}$ , which rounds to 225 Joules as per the provided option.

📊 Diagram / Illustration

✅

B is correct — Using the formula $W = F \cdot s \cdot \cos(\theta)$ , we get $20 \times 13 \times \cos(30°) \approx 225\text{ J}$ .

Core Concepts Used

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Work-Energy Theorem Vector Resolution Trigonometric Functions in Physics

💡 EXAM TIP

Always verify if the angle provided is with the direction of displacement; if the angle is given with the vertical, you must use $\sin(\theta)$ instead.

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