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

The Correct Option is C

Work done (W) equals the dot product of force ($\vec{F}$) and displacement ($\vec{x}$). Given $\vec{F} = (-2\hat{i} + 3\hat{j})$ and $\vec{x} = (3\hat{i})$, W = $(-2\hat{i} + 3\hat{j}) \cdot (3\hat{i}) = -6 \text{ J}$.

The work-energy theorem states that W = ΔKE = KE_f - KE_i. Substituting the calculated work and initial kinetic energy (KE_i = 10 J), we get -6 J = KE_f - 10 J. Therefore, the final kinetic energy (KE_f) is 4 J.