Question

Work done=?

• A. Force $\times$ Distance
• B. Mass $\times$ Velocity
• C. Power $\times$ Time
• D. Force $\times$ Acceleration

Hint:

Always remember: No movement = No work. You can push against a brick wall all day and feel tired, but in the eyes of physics, if the wall didn't move any distance, you did zero Joules of work.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The question asks for the fundamental mathematical relationship or formula used to calculate "Work" in classical mechanics.
Step 2: Key Formula or Approach:
The general physics formula for work is:
\[ W = F \cdot d \cdot \cos(\theta) \]
Where \( F \) is force, \( d \) is displacement (distance), and \( \theta \) is the angle between the force and the direction of motion.
Step 3: Detailed Explanation:

Core Definition: In physics, "work" is done when a force acts upon an object to cause a displacement. If there is no displacement, no physical work is done, regardless of how much force is applied.

Standard Case: When the force is applied in the same direction as the movement (\( \theta = 0^{\circ} \)), the formula simplifies to \( \text{Work} = \text{Force} \times \text{Distance} \). This is the relationship highlighted in option A.

Evaluating Alternatives:
- Mass $\times$ Velocity (B): This is the formula for Momentum (\( p = mv \)).
- Power $\times$ Time (C): While this also equals work (\( W = P \times t \)), in the context of basic mechanics definitions, Force times Distance is the more fundamental definition.
- Force $\times$ Acceleration (D): This does not represent any standard physical quantity; Force is mass times acceleration.

Units of Work: Work is measured in Joules (J). One Joule is the work done when a force of one Newton moves an object through a distance of one meter.

Step 4: Final Answer:
The most direct and fundamental definition of work done provided in the options is the product of Force and Distance.

Answer 2

A concrete example makes the definition of work easy to confirm: imagine pushing a box with a steady force of \(10\text{ N}\) over a distance of \(5\text{ m}\).

1. Force \(\times\) Distance: \(10\text{ N} \times 5\text{ m} = 50\text{ J}\). This is the actual work done on the box, matching the physical definition of energy transferred by pushing it that distance.
2. Mass \(\times\) Velocity: This needs the box's mass and speed, quantities that were never part of the scenario at all, since work depends on force and displacement, not on how heavy the box is or how fast it happens to be moving.
3. Power \(\times\) Time: This would only reproduce the \(50\text{ J}\) figure if you separately knew how long the push took and divided the same \(50\text{ J}\) by that time to get power first, an unnecessary detour when force and distance are already known directly.
4. Force \(\times\) Acceleration: Multiplying \(10\text{ N}\) by an acceleration value produces a quantity with units of \( \text{N}\cdot\text{m/s}^2 \), something that doesn't correspond to energy or work at all, and isn't even a standard physical quantity.

Only multiplying the applied force by the distance moved reproduces the actual energy transferred in the example.

Therefore, the correct answer is Force \(\times\) Distance.