Back to Directory
NEET PHYSICSEasy

A uniform force of (3i^+j^) N(3\hat{i} + \hat{j}) \text{ N} acts on a particle of mass 2 kg2 \text{ kg}. Hence the particle is displaced from position (2i^+k^) m(2\hat{i} + \hat{k}) \text{ m} to position (4i^+3j^−k^) m(4\hat{i} + 3\hat{j} - \hat{k}) \text{ m}. The work done by the force on the particle is:

A

9 J

B

6 J

C

13 J

D

15 J

Step-by-Step Solution

  1. Definition of Work: Work done (WW) by a constant force (F⃗\vec{F}) displacing a particle is the dot product of the force vector and the displacement vector (d⃗\vec{d}). W=F⃗⋅d⃗W = \vec{F} \cdot \vec{d} (Refer to NCERT Class 11, Chapter 6, Section 6.3, Eq. 5.4).
  2. Calculate Displacement Vector (d⃗\vec{d}): Displacement is the change in position vector, d⃗=r⃗f−r⃗i\vec{d} = \vec{r}_f - \vec{r}_i.
  • Initial Position (r⃗i\vec{r}_i): 2i^+0j^+k^2\hat{i} + 0\hat{j} + \hat{k}
  • Final Position (r⃗f\vec{r}_f): 4i^+3j^−k^4\hat{i} + 3\hat{j} - \hat{k} d⃗=(4i^+3j^−k^)−(2i^+0j^+k^)\vec{d} = (4\hat{i} + 3\hat{j} - \hat{k}) - (2\hat{i} + 0\hat{j} + \hat{k}) d⃗=(4−2)i^+(3−0)j^+(−1−1)k^\vec{d} = (4-2)\hat{i} + (3-0)\hat{j} + (-1-1)\hat{k} d⃗=2i^+3j^−2k^ m\vec{d} = 2\hat{i} + 3\hat{j} - 2\hat{k} \text{ m}
  1. Calculate Work Done: Given Force F⃗=3i^+j^=3i^+1j^+0k^ N\vec{F} = 3\hat{i} + \hat{j} = 3\hat{i} + 1\hat{j} + 0\hat{k} \text{ N}. W=F⃗⋅d⃗=(3)(2)+(1)(3)+(0)(−2)W = \vec{F} \cdot \vec{d} = (3)(2) + (1)(3) + (0)(-2) W=6+3+0=9 JW = 6 + 3 + 0 = 9 \text{ J} (Refer to NCERT Class 11, Section 6.1.1 for Scalar Product rules).
  2. Note on Mass: The mass of the particle (2 kg2 \text{ kg}) is extraneous information for calculating work done in this context, as work depends only on force and displacement.
Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started