Q103 / 200 · NEET 2024 (Re-NEET) · Physics

An object moving along horizontal $x$-direction with kinetic energy 10 J is displaced through $x=(3 \hat{i}) \mathrm{m}$ by the force $\vec{F}=(-2 \hat{i}+3 \hat{j}) \mathrm{N}$. The kinetic energy of the object at the end of the displacement $x$ is

  • A
    10 J
  • B
    16 J
  • Correct answer
    C
    4 J
  • D
    6 J

Solution

Work energy theorem, $W_{\mathrm{al}}=\Delta \mathrm{K} . \mathrm{E}\left(W_{\mathrm{al}}=\right.$ work done by all forces $)$ $\Rightarrow K_f-K_l=\vec{F} \cdot \Delta \overline{\mathrm{x}}$ $\Rightarrow K_f-10=(-2 \hat{i}+3 \hat{j}) \cdot(3 \hat{i})$ $K_f-10=-6$ $K_{\mathrm{f}}=4 \mathrm{~J}$

Hence, the correct answer is option: 'C'.

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