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An object moving along horizontal x-direction with kinetic energy 10 J is displaced through $x=(3\hat{i})$ 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
A
10 J
B
16 J
C
4 J
D
6 J
Detailed Solution
By the work-energy theorem, $W=\Delta KE$. $W=\vec{F}\cdot\vec{x}=(-2\hat{i}+3\hat{j})\cdot(3\hat{i})=-6$ J. So $K_f-10=-6\Rightarrow K_f=4$ J.