Question

The work done by the force \[\overset{\to }{\mathop{F}}\,=\overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\overset{\to }{\mathop{k}}\,\] acting on a particle is displaced from A (3, 3, 3) to the point B (4, 4, 4) is,
(a) 2 units
(b) 3 units
(c) 4 units
(d) 7 units

Answer 1

Hint: First calculate the distance by using the formula \[\overset{\to }{\mathop{AB}}\,=\overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{a}}\,\] and then use the formula \[W=\overset{\to }{\mathop{F}}\,.\overset{\to }{\mathop{AB}}\,\] to get the value of work done. Use the formula \[\overset{\to }{\mathop{p}}\,.\overset{\to }{\mathop{q}}\,=\left( \overset{\to }{\mathop{ai}}\,+\overset{\to }{\mathop{bj}}\,+\overset{\to }{\mathop{ck}}\, \right).\left( \overset{\to }{\mathop{di}}\,+\overset{\to }{\mathop{ej}}\,+\overset{\to }{\mathop{fk}}\, \right)=ad+be+cf\] to calculate the dot product.

Complete step-by-step answer:
To find out the work done by the given force we should write the given equation first, therefore,
\[\overset{\to }{\mathop{F}}\,=\overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\overset{\to }{\mathop{k}}\,\] ……………………………………….. (1)
Also, A (3, 3, 3) and B (4, 4, 4) ……………………………………… (2)
As we have given the two points A and B therefore it’s position vector will become,
\[\overset{\to }{\mathop{a}}\,=3\overset{\to }{\mathop{i}}\,+3\overset{\to }{\mathop{j}}\,+3\overset{\to }{\mathop{k}}\,\] and \[\overset{\to }{\mathop{b}}\,=4\overset{\to }{\mathop{i}}\,+4\overset{\to }{\mathop{j}}\,+4\overset{\to }{\mathop{k}}\,\] ……………………………………… (3)
The distance between two points is given by the formula, \[\overset{\to }{\mathop{AB}}\,=\overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{a}}\,\] therefore,
\[\overset{\to }{\mathop{AB}}\,=\overset{\to }{\mathop{b}}\,-\overset{\to }{\mathop{a}}\,\]
If we put the values of equation (3) in the above formula we will get,
\[\therefore \overset{\to }{\mathop{AB}}\,=\left( 4\overset{\to }{\mathop{i}}\,+4\overset{\to }{\mathop{j}}\,+4\overset{\to }{\mathop{k}}\, \right)-\left( 3\overset{\to }{\mathop{i}}\,+3\overset{\to }{\mathop{j}}\,+3\overset{\to }{\mathop{k}}\, \right)\]
Further simplification in the above equation will give,
\[\therefore \overset{\to }{\mathop{AB}}\,=4\overset{\to }{\mathop{i}}\,+4\overset{\to }{\mathop{j}}\,+4\overset{\to }{\mathop{k}}\,-3\overset{\to }{\mathop{i}}\,-3\overset{\to }{\mathop{j}}\,-3\overset{\to }{\mathop{k}}\,\]
By rearranging the above equation we will get,
\[\therefore \overset{\to }{\mathop{AB}}\,=\left( 4\overset{\to }{\mathop{i}}\,-3\overset{\to }{\mathop{i}}\, \right)+\left( 4\overset{\to }{\mathop{j}}\,-3\overset{\to }{\mathop{j}}\, \right)+\left( 4\overset{\to }{\mathop{k}}\,-3\overset{\to }{\mathop{k}}\, \right)\]
\[\therefore \overset{\to }{\mathop{AB}}\,=\overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\overset{\to }{\mathop{k}}\,\] ………………………………………………………. (4)
To proceed further in the solution we should know the formula of work done given below,
Formula:
\[W=\overset{\to }{\mathop{F}}\,.\overset{\to }{\mathop{d}}\,\] Where \[\overset{\to }{\mathop{d}}\,\] is the distance covered which is \[\overset{\to }{\mathop{AB}}\,\] in this case,
Therefore the work by the given force is given by,
\[W=\overset{\to }{\mathop{F}}\,.\overset{\to }{\mathop{AB}}\,\]
If we put the values of equation (1) and equation (4) in the above equation we will get,
\[W=\left( \overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\overset{\to }{\mathop{k}}\, \right).\left( \overset{\to }{\mathop{i}}\,+\overset{\to }{\mathop{j}}\,+\overset{\to }{\mathop{k}}\, \right)\]
Now to proceed further in the solution we should know the the formula given below,
Formula:
If \[\overset{\to }{\mathop{p}}\,=\overset{\to }{\mathop{ai}}\,+\overset{\to }{\mathop{bj}}\,+\overset{\to }{\mathop{ck}}\,\] and \[\overset{\to }{\mathop{q}}\,=\overset{\to }{\mathop{di}}\,+\overset{\to }{\mathop{ej}}\,+\overset{\to }{\mathop{fk}}\,\] then their dot product is given by, \[\overset{\to }{\mathop{p}}\,.\overset{\to }{\mathop{q}}\,=\left( \overset{\to }{\mathop{ai}}\,+\overset{\to }{\mathop{bj}}\,+\overset{\to }{\mathop{ck}}\, \right).\left( \overset{\to }{\mathop{di}}\,+\overset{\to }{\mathop{ej}}\,+\overset{\to }{\mathop{fk}}\, \right)=ad+be+cf\].
By using the above formula in ‘W’ we will get,
\[\therefore W=1\times 1+1\times 1+1\times 1\]
\[\therefore W=1+1+1\]
Therefore, W = 3 units.
Therefore the work done by a given force from point A to point B is equal to 3 units.
Therefore the correct answer is option (b).

Note: You can also solve this problem by calculating \[\left| \overset{\to }{\mathop{F}}\, \right|\] and distance AB by using distance formula and then using the formula \[W=\left| \overset{\to }{\mathop{F}}\, \right|.AB\] to get the quick answer.