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NEET PHYSICSWORK, ENERGY AND POWEREasy

Question

A uniform force of (3i^+j^)(3\hat{i} + \hat{j}) N acts on a particle of mass 2 kg. The particle is displaced from the position (2i^+k^)(2\hat{i} + \hat{k}) m to the position (4i^+3j^k^)(4\hat{i} + 3\hat{j} - \hat{k}) m. The work done by the force on the particle is:

A

6 J

B

13 J

C

15 J

D

9 J

Step-by-Step Solution

According to the definition of work done by a constant force, work (WW) is the scalar product of the force vector (F\mathbf{F}) and the displacement vector (d\mathbf{d}), expressed as W=FdW = \mathbf{F} \cdot \mathbf{d} .

  1. Find the displacement vector (d\mathbf{d}): Displacement is the difference between the final position (r2\mathbf{r_2}) and the initial position (r1\mathbf{r_1}) . d=r2r1=(4i^+3j^k^)(2i^+k^)\mathbf{d} = \mathbf{r_2} - \mathbf{r_1} = (4\hat{i} + 3\hat{j} - \hat{k}) - (2\hat{i} + \hat{k}) d=(42)i^+(30)j^+(11)k^=2i^+3j^2k^ m\mathbf{d} = (4-2)\hat{i} + (3-0)\hat{j} + (-1-1)\hat{k} = 2\hat{i} + 3\hat{j} - 2\hat{k} \text{ m}

  2. Calculate the scalar product: Multiply the corresponding components of the force F=(3i^+j^+0k^)\mathbf{F} = (3\hat{i} + \hat{j} + 0\hat{k}) N and the displacement vector . W=(3)(2)+(1)(3)+(0)(2)W = (3)(2) + (1)(3) + (0)(-2) W=6+3+0=9 JW = 6 + 3 + 0 = 9 \text{ J}

Thus, the work done by the force is 9 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 POWERuniformparticleparticledisplacedposition

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