An object moving along horizontal -direction with kinetic energy 10 J is displaced through , by the force . The kinetic energy of the object at the end of the displacement is: [2024]

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Solution

Q.
An object moving along horizontal $x$ -direction with kinetic energy 10 J is displaced through $\vec{x} = (3 \hat{i})$ $m$ , by the force $\vec{F} = (- 2 \hat{i} + 3 \hat{j})$ $N$ . The kinetic energy of the object at the end of the displacement $x$ is: [2024]

1 10 J

2 16 J

3 4 J

4 6 J

Ans.
(3)

Given, $\vec{F} = (- 2 \hat{i} + 3 \hat{j}) N$ or $\vec{x} = (3 \hat{i}) m$

Initial Kinetic energy, $(K_{i}) = 10 J$

$W = \vec{F} \cdot \vec{x} = (- 2 \hat{i} + 3 \hat{j}) \cdot (3 \hat{i}) = - 6 J$

Using work-energy theorem, $W = Δ K . E .$

$- 6 = K_{f} - K_{i} or - 6 = K_{f} - 10;$ $K_{f} = 4 J$

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