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