Correct Answer: Option C

Explanation

F = 6N, m = 2kg, t = 4s, u = 0 (stationary or at rest)

K.E = \(\frac{1}{2}\)mv\(^2\)

F = \(\frac{\text{mv}}{\text{t}}\)

from rest

K.E = \(\frac{1}{2}m(\frac{\text{Ft}}{\text{m}})^2\) = \(\frac{1}{2}\) x 2 x (\(\frac{6 \times 4}{2})^2\) = 12\(^2\) = 144J.

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