Question:medium
If a force of $ 10 \, \text{N} $ displaces a body by $ 5 \, \text{m} $ in the direction of the force, the work done is:
If a force of $ 10 \, \text{N} $ displaces a body by $ 5 \, \text{m} $ in the direction of the force, the work done is:
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Key Fact: Work \( = Fd\cos\theta \); maximum when \( \theta = 0^\circ \)
Updated On: Jan 13, 2026
- \( 2 \, \text{J} \)
- \( 50 \, \text{J} \)
- \( 0 \, \text{J} \)
- \( 5 \, \text{J} \)
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The Correct Option is B
Solution and Explanation
The task involves determining the work performed when a \(10 \, \text{N}\) force causes an object to move \(5 \, \text{m}\). The formula for work done (\(W\)) is:
\(W = F \times d \times \cos(\theta)\)
In this formula:
- \(F\) represents the applied force;
- \(d\) represents the displacement;
- \(\theta\) is the angle between the force vector and the displacement vector.
Given that the force and displacement are aligned, \(\theta = 0^\circ\), which means \(\cos(0^\circ) = 1\). This simplifies the work calculation to:
\(W = 10 \, \text{N} \times 5 \, \text{m} \times 1\)
\(W = 50 \, \text{J}\)
Therefore, the total work done is \(50 \, \text{J}\).
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