Question

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

• A. $10 \, \text{J}$
• B. $16 \, \text{J}$
• C. $4 \, \text{J}$
• D. $6 \, \text{J}$

Answer 1

Answer: (C)

$4 \, \text{J}$

Answer 2

Work done (W) = $\vec{F} \times \vec{x} = (-2\hat{i} + 3\hat{j}) \times (3\hat{i}) = -6 \text{ J}$.

Work-energy theorem : W = ΔKE = KEf - KEi = -6 J = KEf - 10 J KEf = 4 J