What is the work done if a constant force of 50 N is applied a... | General Science MCQ | Examoogle

General SciencePhysics

What is the work done if a constant force of 50 N is applied at an angle of 60 degrees to the direction of displacement of 10 m?

250 J

500 J

433 J

200 J

Correct Answer

🔬 SC • Science Concept & PrincipleGeneral SciencePhysics

Option A

250 J

Quick Summary:

Work done is defined as the dot product of the force vector and the displacement vector. When a force is applied at an angle, only the component of the force acting in the direction of displacement contributes to the work done.

🎯

Crack Physics with Examoogle SuperTeachers

Get Free Demo Classes & Live Mocks →

🔬SCScience SolutionConcept & Principle

💡 Explanation

💡 Everyday Analogy (Real-World Intuition)

Think of pulling a suitcase with wheels; if you pull it at an angle, only the horizontal part of your pull moves it forward, while the upward part just makes the suitcase slightly lighter.

🧠 Memory Mnemonic / Shortcut Aid

F-S-Cos: Force times displacement times the cosine of the angle.

🔢 Key Formulas

$W = Fs \cos(\theta)$ — General formula for work done

$\cos(60°) = \frac{1}{2} = 0.5$ — Trigonometric constant for this specific angle

⚙️ Working Principle

The work done $W$ is calculated using the formula $W = F \cdot s \cdot \cos(\theta)$ , where $F$ is the magnitude of force, $s$ is displacement, and $\theta$ is the angle between them. Given $F = 50\text{ N}$ , $s = 10\text{ m}$ , and $\theta = 60°$ , we have $W = 50 \times 10 \times \cos(60°)$ . Since $\cos(60°) = 0.5$ , the work done is $500 \times 0.5 = 250\text{ J}$ .

📌 Key Points

▸
Work is a scalar quantity, meaning it has magnitude but no direction.
▸
The SI unit of work is the Joule ( $\text{J}$ ), which is equivalent to $1 \text{ N} \cdot \text{m}$ .
▸
If $\theta = 90°$ , the work done is zero because $\cos(90°) = 0$ .

✅ Advantages

▸
Provides a clear method to quantify energy transfer through mechanical action.

❌ Disadvantages / Limitations

▸
Does not account for non-constant forces without the use of calculus (integration).

🛠️ Applications / Uses

▸
Calculating energy expenditure for moving objects.
▸
Designing mechanical systems like levers and pulleys.

📄 Additional Information

▸
The value of $\cos(0°) = 1$ (maximum work) and $\cos(180°) = -1$ (negative work/opposing motion).
▸
Option B (500 J) incorrectly assumes the force is applied entirely parallel to the displacement ( $\theta = 0°$ ).
▸
Option C (433 J) is incorrect because it likely uses $\sin(60°) \approx 0.866$ instead of $\cos(60°)$ .

📊 Diagram / Illustration

✅

A is correct — The work done is calculated as $50 \times 10 \times \cos(60°) = 250\text{ J}$ .

Core Concepts Used

Click any tag to open in AI Tutor

Work-Energy Theorem Vector Dot Product Trigonometric Components of Force

💡 EXAM TIP

Always verify if the angle provided is with the direction of displacement or with the normal (vertical) to the surface; if it is with the normal, use $\sin$ instead of $\cos$ .

Was this solution helpful?

⚡ Practice on Android App · 4.8 ★

Solve 10,000+ More Questions with AI Tutor

Get instant step-by-step AI explanations, full shift Previous Year Papers (SSC GD, RRB, SSC JE), and regional rank analysis on the official Examoogle App.

Free Download • 100K+ Aspirants

Discussion (0)

Loading discussion...

Prev Next

Correct Answer

Solve 10,000+ More Questions with AI Tutor

Related Questions

Discussion (0)