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NEET 2026
Exam
NEET
Year
2026
Subject
Physics
Chapter
Chapter

NEET 2026 · Physics · Chapter

A particle is moving in the x - y plane with an initial kinetic...

A particle is moving in the xx-yy plane with an initial kinetic energy of 20 J20\text{ J}. A constant force F⃗=(3i^+4j^) N\vec{F} = (3\hat{i} + 4\hat{j})\text{ N} acts on the particle as it moves from the coordinates (1 m,2 m)(1\text{ m}, 2\text{ m}) to (3 m,−1 m)(3\text{ m}, -1\text{ m}). The kinetic energy of the particle at the final position is:

Options

  1. A14 J14\text{ J}
  2. B26 J26\text{ J}
  3. C38 J38\text{ J}
  4. D6 J6\text{ J}

Correct answer

A. 14 J14\text{ J}

Step-by-step solution

Initial position vector r⃗i=i^+2j^\vec{r}_i = \hat{i} + 2\hat{j}

Final position vector r⃗f=3i^−j^\vec{r}_f = 3\hat{i} - \hat{j}

Displacement vector Δr⃗=r⃗f−r⃗i=(3−1)i^+(−1−2)j^=2i^−3j^\Delta\vec{r} = \vec{r}_f - \vec{r}_i = (3 - 1)\hat{i} + (-1 - 2)\hat{j} = 2\hat{i} - 3\hat{j}

Work done by the constant force W=F⃗⋅Δr⃗=(3i^+4j^)⋅(2i^−3j^)W = \vec{F} \cdot \Delta\vec{r} = (3\hat{i} + 4\hat{j}) \cdot (2\hat{i} - 3\hat{j})

W=(3)(2)+(4)(−3)=6−12=−6 JW = (3)(2) + (4)(-3) = 6 - 12 = -6\text{ J}

Using the work-energy theorem, W=Kf−KiW = K_f - K_i

−6=Kf−20-6 = K_f - 20

Kf=14 JK_f = 14\text{ J}

Answer: 14 J14\text{ J}
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