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
AB is a part of an electrical circuit (see figure). The potential difference \(V_A - V_B\), at the instant when current \(i = 2\) A and is increasing at a rate of 1 amp/second is: