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NEET PHYSICSWORK, ENERGY AND POWEREasy

Question

A particle moves from position r1=3i^+2j^6k^\vec{r}_1= 3\hat{i}+2\hat{j}-6\hat{k} to position r2=14i^+13j^+9k^\vec{r}_2=14\hat{i}+13\hat{j}+9\hat{k} under the action of force 4i^+j^+3k^ N4\hat{i}+\hat{j}+3\hat{k} \text{ N}. The work done will be:

A

100 J100 \text{ J}

B

50 J50 \text{ J}

C

200 J200 \text{ J}

D

75 J75 \text{ J}

Step-by-Step Solution

Given, Initial position vector, r1=3i^+2j^6k^\vec{r}_1 = 3\hat{i} + 2\hat{j} - 6\hat{k} Final position vector, r2=14i^+13j^+9k^\vec{r}_2 = 14\hat{i} + 13\hat{j} + 9\hat{k} Force acting on the particle, F=4i^+j^+3k^ N\vec{F} = 4\hat{i} + \hat{j} + 3\hat{k} \text{ N}

The displacement vector d\vec{d} is given by: d=r2r1\vec{d} = \vec{r}_2 - \vec{r}_1 d=(14i^+13j^+9k^)(3i^+2j^6k^)\vec{d} = (14\hat{i} + 13\hat{j} + 9\hat{k}) - (3\hat{i} + 2\hat{j} - 6\hat{k}) d=(143)i^+(132)j^+(9+6)k^=11i^+11j^+15k^\vec{d} = (14 - 3)\hat{i} + (13 - 2)\hat{j} + (9 + 6)\hat{k} = 11\hat{i} + 11\hat{j} + 15\hat{k}

The work done WW by a constant force is the dot product of the force vector and the displacement vector : W=FdW = \vec{F} \cdot \vec{d} W=(4i^+j^+3k^)(11i^+11j^+15k^)W = (4\hat{i} + \hat{j} + 3\hat{k}) \cdot (11\hat{i} + 11\hat{j} + 15\hat{k}) W=(4×11)+(1×11)+(3×15)W = (4 \times 11) + (1 \times 11) + (3 \times 15) W=44+11+45=100 JW = 44 + 11 + 45 = 100 \text{ J}

Thus, the work done is 100 J100 \text{ J}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WORK, ENERGY AND POWER. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWORK, ENERGY AND POWERparticlepositionhatihatjhatkpositionvecrhatihatjhatk

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