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A force F= 2i +3j- 4kN acts on a particle which is constrained to move in the XOY plane along the line x = y. If the particle moves 5 $\sqrt{2}$ m, the work done by force in joule is

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$5 \sqrt{58}$

25 $\sqrt{2}$

answer is C.

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Detailed Solution

F = 2 i + 3 j - 4 k as x=y s=5 $\sqrt{2}$ then s = 5 i + 5 j

work done = 2x5 + 3x 5 = 25 joule

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