Question

Forces acting on a particle have magnitudes 5,3,1$kg.wt$ and act in the directions of the vectors $6i + 2j + 3k$ , $3i - 2j + 6k$ and $2i - 3j - 6k$ 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

Answer 1

Hint: The direction of the forces acting on a particle can be known by making the use of vectors. Therefore vectors are the quantities that have both magnitude and direction. It is basically represented by using the symbol ‘hat or ‘cap.

Formula Used:
Work done $W = {\overrightarrow F _{net}}.\overrightarrow d $
Complete step by step answer:
Step I:
Formula to find unit vector $ = \dfrac{{Vector}}{{Vector\;magnitude}}$
If ‘a’ is any vector then, formula will be
$\widehat a = \dfrac{a}{{\left| a \right|}}$
Where $\left| a \right|$ is the magnitude of the vector. Its formula is
$\left| a \right| = \sqrt {{x^2} + {y^2} + {z^2}} $
Step II:
The unit vectors in the direction of the given vectors will be
\[\widehat {{F_1}} = \] $\dfrac{{6\widehat i + 2\widehat j + 3\widehat k}}{{\sqrt {{{(6)}^2} + {{(2)}^2} + {{(3)}^2}} }}$ \[ = \dfrac{{6\widehat i + 2\widehat j + 3\widehat k}}{{\sqrt {36 + 4 + 9} }} = \dfrac{{6\widehat i + 2\widehat j + 3\widehat k}}{{\sqrt {49} }}\] $ = \dfrac{{6\widehat i + 2\widehat j + 3\widehat k}}{7}$
Similarly $\widehat {{F_2}} = \dfrac{{3\widehat i - 2\widehat j + 6\widehat k}}{{\sqrt {49} }} = \dfrac{{3\widehat i - 2\widehat j + 6\widehat k}}{7}$
$\widehat {{F_3}} = \dfrac{{2\widehat i - 3\widehat j - 6\widehat k}}{{\sqrt {49} }} = \dfrac{{2\widehat i - 3\widehat j - 6\widehat k}}{7}$
Step III:
Given ${F_1},{F_2},{F_3}$ are forces of magnitude $5,3,1$
${F_1} = 5 \times \dfrac{{6\widehat i + 2\widehat j + 3\widehat k}}{7}$
${F_2} = 3 \times \dfrac{{3\widehat i - 2\widehat j + 6\widehat k}}{7}$
\[{F_3} = 1 \times \dfrac{{2\widehat i - 3\widehat j - 6\widehat k}}{7}\]
Step IV:
Resultant force is the sum of individual forces.
$F = {F_1} + {F_2} + {F_3}$
Substituting values and solving
$F = \dfrac{1}{7}[\{ 5(6\widehat i + 2\widehat j + 3\widehat k)\} + \{ 3(3\widehat i - 2\widehat j + 6\widehat k)\} + \{ 1(2\widehat i - 3\widehat j - 6\widehat k)\} ]$
\[F = \dfrac{1}{7}[30\widehat i + 10\widehat j + 15\widehat k + 9\widehat i - 6\widehat j + 18\widehat k + 2\widehat i - 3\widehat j - 6\widehat k]\]
$F = \dfrac{1}{7}(41\widehat i + \widehat j + 27\widehat k)$
Given that the Body is displaced from (2,-1,-3) to (5,-1,1). So the displacement vector will be
$\overrightarrow d = 3\overrightarrow i + 4\overrightarrow k $
Work done $W = {\overrightarrow F _{net}}.\overrightarrow d $
Or $\dfrac{1}{7}(41\widehat i + \widehat j + 27\widehat k).(3\widehat i + 4\widehat k)$
Only like vectors are to be multiplied. That is $\widehat i.\widehat i = \widehat j.\widehat j = \widehat k.\widehat k = 1$
And $\widehat i.\widehat j = \widehat j.\widehat k = \widehat k.\widehat i = 0$
Therefore, Work done is
=$\dfrac{1}{7}(41 \times 3 + 27 \times 4)$
 $W = \dfrac{{231}}{7} = 33units$

The work done by the resultant forces in moving the object from one place to another is $33units$.
Therefore, Option B is the right answer.

Note: A vector that has a magnitude 1 is called as a unit vector. Any vector can be changed or converted into a unit vector, if the given vector is divided by its magnitude. The resultant motion of the particle or the object is known by making the use of vectors. Also note, kg-wt and N are different units for expressing force, 1kg-wt = 9.8 N.