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Instantaneous Velocity

Forces acting...

Question

Forces acting on a particle have magnitudes of $5, 3$ and $1$ units and act in the direction of the vectors $6 i + 2 j + 3 k, 3 i - 2 j + 6 k$ and $2 i - 3 j - 6 k,$ respectively. They remain constant while the particle is displaced from the point $A (2, - 1, - 3)$ to $B (5, - 1, 1)$ . The work done is equal to

31

units

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33

units

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34

units

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44

units

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Solution

The correct option is B $33$ units
The unit vectors in the direction of the given vectors are
$\frac{6 i + 2 j + 3 k}{\sqrt{36 + 4 + 9}} = \frac{6 i + 2 j + 3 k}{7}, \frac{3 i - 2 j + 6 k}{7}$
and $\frac{2 i - 3 j - 6 k}{7}$
Since $F_{1}, F_{2}$ and $F_{3}$ are the forces of magnitude $5, 3$ and $1$ units, we have
$F_{1} = \frac{5}{7} (6 i + 2 j + 3 k),$
$F_{2} = \frac{3}{7} (3 i - 2 j + 6 k),$
$F_{3} = \frac{1}{7} (2 i - 3 j - 6 k),$
Therefore the resultant is
$R = F_{1} + F_{2} + F_{3} = \frac{1}{7} (41 i + j + 27 k)$ and $A B = 3 i + 4 k .$
Hence the work done is $R \cdot A B = \frac{1}{7} (41 \times 3 + 27 \times 4) = \frac{231}{7} = 33$ units.

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