A particle acted on by constant forces $3 i + j - k$ and $4 i + j - 3 k$ is displaced form the point $i + 2 j + 3 k$ to $5 i + 4 j + k .$ Find the total work done by the forces.

22 units

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23 units

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24 units

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25 units

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Solution

The correct option is C 24 units
We know that the work done by several forces is equal to the work done by their resultant
$R = F_{1} + F_{2} = (4 i + j - 3 k) + (3 i + j - k) = 7 i + 2 j - 4 k .$
Again if the point is displaced from $P$ to $Q$ , then $\to P Q = P . V$ of $Q - P . V$ of $P$
$= (5 i + 4 j + k) - (i + 2 j + 3 k) = 4 i + 2 j - 2 k$
Hence the work done $= R . \to P Q$ $= (7 i + 2 j + 4 k) . (4 i + 2 j - 2 k) = 28 + 4 - 8 = 24$ units.

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