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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Work and Energy

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Work and Energy

A particle travels from a position r _1 = (2 m) i + (3 m) j to a...

A particle travels from a position r⃗1=(2 m)i^+(3 m)j^\vec{r}_1 = (2 \text{ m})\hat{i} + (3 \text{ m})\hat{j} to a new position r⃗2=(3 m)i^+(2 m)j^\vec{r}_2 = (3 \text{ m})\hat{i} + (2 \text{ m})\hat{j} while experiencing a force F⃗=(5 N)i^+(5 N)j^\vec{F} = (5 \text{ N})\hat{i} + (5 \text{ N})\hat{j}. Determine the work done by this force on the particle over the displacement.

Options

  1. A10 J10 \text{ J}
  2. B25 J25 \text{ J}
  3. C50 J50 \text{ J}
  4. D0 J0 \text{ J}

Correct answer

D. 0 J0 \text{ J}

Step-by-step solution

The displacement vector of the particle is given by:

Δr⃗=r⃗2−r⃗1\Delta\vec{r} = \vec{r}_2 - \vec{r}_1

Δr⃗=(3i^+2j^)−(2i^+3j^)\Delta\vec{r} = (3\hat{i} + 2\hat{j}) - (2\hat{i} + 3\hat{j})

Δr⃗=i^−j^ m\Delta\vec{r} = \hat{i} - \hat{j} \text{ m}

The work done by the constant force is the dot product of the force and the displacement:

W=F⃗⋅Δr⃗W = \vec{F} \cdot \Delta\vec{r}

W=(5i^+5j^)⋅(i^−j^)W = (5\hat{i} + 5\hat{j}) \cdot (\hat{i} - \hat{j})

W=(5)(1)+(5)(−1)W = (5)(1) + (5)(-1)

W=5−5=0 JW = 5 - 5 = 0 \text{ J}
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