A body of mass 2kg starts moving uniformly from the point (1,2,3) to point (2,3,5) in 3 seconds. The net work done during 3 seconds is
A
34J
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B
23J
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C
49J
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D
916J
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Solution
The correct option is B23J
Position vector →S is given by →S=(2−1)^i+(3−2)^j+(5−3)^k =^i+^j+2^k Velocity vector →v obtained by the dividing the position vector by the time taken Final velocity →v=^i+^j+2^k3 Initial velocity =0 Now, According to work energy theorem Net work done = Change in KE Net work done =12m(→v.→v) ⇒12×2×(19×(1×1+1×1+2×2)) =69J=23J