Question

Forces acting on a particle have magnitudes $$5, 3, 1 \ kg.wt$$ and act in the directions of the vectors $$\displaystyle 6i+2j+3k,3i-2j+6k$$ and $$\displaystyle 2i-3j-6k$$ respectively.These remain constant while the particle is displaced from A (4, -2, -6) to B (7,-2,-2). Find the work done.

A
21 units.
B
33 units.
C
47 units.
D
52 units.

Solution

The correct option is C $$33$$ units.Unit vectors in the direction of the forces are $$\displaystyle \frac{6i+2j+3k}{\sqrt{\left ( 36+4+9 \right )}},\frac{3i-2j+6k}{7},\frac{2i-3j-6k}{7}$$Hence if $$\displaystyle F_{1}, F_{2}, F_{3},$$ be the forces of magnitudes 5, 3, 1, then they are$$\displaystyle \frac{5}{7}\left ( 6i+2j+3k \right ),\frac{3}{7}\left ( 3i-2j+6k \right )$$ and $$\displaystyle \frac{1}{7}\left ( 2i-3j-6k \right ).$$ Again if $$R$$ be their resultant then$$\displaystyle R= F_{1}+F_{2}+F_{3}.$$    $$\displaystyle \therefore R= \frac{1}{7}\left [ \left ( 30+9+2 \right )i+\left ( 10-6-3 \right )j+\left ( 15+18-6 \right )k \right ]$$ $$\displaystyle = \frac{1}{7}\left ( 41i+j+27k \right ).$$ $$\displaystyle = \frac{1}{7}\left ( 41i+j+27k \right ).$$Again $$\displaystyle \vec{AB}= P.V$$ of $$\displaystyle B \ - \ P.V$$ A $$\displaystyle = \left ( 4i-2j-6k \right ) - \left ( 7i-2j-2k \right )= -3i+0j-4k$$The work done by  various forces is equal to work done by their resultant.$$\displaystyle \therefore$$ Work done $$\displaystyle = R.\vec{AB}$$ $$\displaystyle = \frac{1}{7}\left [ 41i+j+27k\right ]. \left [ -3i-4k \right ]$$ $$\displaystyle = \frac{1}{7}\left [ -123+0-108 \right ]= \frac{-231}{7}= -33$$So work done is $$|-33| = 33$$ units.Maths

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