Question

A particle moves from position $3\hat i + 2\hat j - 6\hat k$ to $14\hat i + 13\hat j + 9\hat k$ due to a force $F = \left( {4\hat i + \hat j + 3\hat k} \right)\,N$. If the displacement is in centimetre, then the work done will be:
(A) $1\,J$
(B) $2\,J$
(C) $3\,J$
(D) $2.5\,J$

Answer 1

The work done is the product of the force vector to the distance vector. But here the two distance vectors are given, so the difference of the two-distance vector is taken as the distance vectors, then the work done by the force on the particle is determined.
Useful formula
The work done by the force vector is given by,
$W = \vec F.\vec r$
Where, $W$ is the work done by the force on the particle, $\vec F$ is the force vector and $\vec r$ is the distance vector.

Complete step by step solution
Given that,
The initial position of the particle is given by, ${\vec r_1} = 3\hat i + 2\hat j - 6\hat k$
The final position of the particle is given by, ${\vec r_2} = 14\hat i + 13\hat j + 9\hat k$
The force vector is given by, $F = \left( {4\hat i + \hat j + 3\hat k} \right)\,N$
The distance of the particle moves can be determined by,
$\vec r = {\vec r_2} - {\vec r_1}$
By substituting the ${\vec r_1}$ and ${\vec r_2}$ in the above equation, then the above equation is written as,
$\vec r = \left( {14\hat i + 13\hat j + 9\hat k} \right) - \left( {3\hat i + 2\hat j - 6\hat k} \right)$
By rearranging the terms in the above equation, then the above equation is written as,
$\vec r = 14\hat i + 13\hat j + 9\hat k - 3\hat i - 2\hat j + 6\hat k$
By subtracting the terms in the above equation, then the above equation is written as,
$\vec r = 11\hat i + 11\hat j + 15\hat k$
Now, the work done is given by,
$W = \vec F.\vec r$
By substituting the force vector and the distance vector in the above equation, then the above equation is written as,
$W = \left( {4\hat i + \hat j + 3\hat k} \right).\left( {11\hat i + 11\hat j + 15\hat k} \right)$
By multiplying the terms in the above equation, then the above equation is written as,
$W = \left( {44 + 11 + 45} \right) \times {10^{ - 2}}$
By adding the terms in the above equation, then the above equation is written as,
$W = 100 \times {10^{ - 2}}$
Then the above equation is also written as,
$W = 1\,J$

Hence, the option (A) is the correct answer.

Note While subtracting the two vector equations, the coefficient of the $\hat i$ is subtracted with the coefficient of the $\hat i$ only, like that the coefficient of the $\hat j$ and $\hat k$ also subtracted with the coefficient of the $\hat j$ and $\hat k$. The same condition is also used in the multiplication of the two-vector equation also. In the work done formula the term ${10^{ - 2}}$ is included because in the question it is given that displacement is centimetre.