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A force of 10i - 3j + 6 k newton acts on a body of 5kg and it displaces from 6i+5j-3k metre to 10 i -2j+7k metre. The work done is

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100 j

121 j

None

answer is C.

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Detailed Solution

$\begin{array}{l} W = F \cdot ds = (10 i - 3 j + 6 k) \cdot \\ [(10 i - 2 j + 7 k) - (6 i + 5 j - 3 k)]] \\ = (10 i - 3 j + 6 k) \cdot (4 i - 7 j + 10 k) \\ = 40 + 21 + 60 = 121 J \end{array}$

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