Question

A particle of mass $ 0.5Kg $ is displaced from position $ {\vec r_1}\left( {2,3,1} \right) $ to $ {\vec r_2}\left( {4,3,2} \right) $ by applying a force of magnitude $ 30N $ which is acting along $ \left( {\hat i + \hat j + \hat k} \right) $ . The work done by the force is
 $ \left( A \right)10\sqrt 3 J \\
  \left( B \right)30\sqrt 3 J \\
  \left( C \right)30J \\
  \left( D \right)none\ of\ these \\ $

Answer 1

Hint :In order to solve this question, we are going to firstly find the length of the position vector from $ {\vec r_1}\left( {2,3,1} \right) $ to $ {\vec r_2}\left( {4,3,2} \right) $ , then the force vector along the direction of $ \left( {\hat i + \hat j + \hat k} \right) $ is found by multiplying the magnitude with the unit vector of $ \left( {\hat i + \hat j + \hat k} \right) $ and thus, the work done is calculated.
The position vector of a particle from position $ {\vec r_1} $ to $ {\vec r_2} $ is given by
 $ {\vec r_{12}} = {\vec r_2} - {\vec r_1} $
The work done by a force $ \vec F $ to cause a displacement of $ {\vec r_{12}} $ is
 $ W = \vec F \cdot {\vec r_{12}} $

Complete Step By Step Answer:
It is given in this question, that the particle of mass $ 0.5Kg $ , is displaced from position $ {\vec r_1}\left( {2,3,1} \right) $ to $ {\vec r_2}\left( {4,3,2} \right) $ , thus, the position vector for this particle will be
 $ {\vec r_{12}} = {\vec r_2} - {\vec r_1} $
Thus, putting values to find the position vector
 $ \Rightarrow {{\vec r}_{12}} = \left( {4 - 2} \right)\hat i + \left( {3 - 3} \right)\hat j + \left( {2 - 1} \right)\hat k \\
   \Rightarrow {{\vec r}_{12}} = 2\hat i + \hat k \\ $
Now as it is given that a force of the magnitude $ 30N $ is acting along the direction of the vector $ \left( {\hat i + \hat j + \hat k} \right) $
Thus, the force vector can be written as
 $ \vec F = 30 \times \dfrac{{\left( {\hat i + \hat j + \hat k} \right)}}{{\sqrt 3 }} = 10\sqrt 3 \left( {\hat i + \hat j + \hat k} \right) $
Now, as we know that the work done is given by the dot product of the force and the displacement vector, so,
 $ W = \vec F \cdot {{\vec r}_{12}} = \left( {10\sqrt 3 \left( {\hat i + \hat j + \hat k} \right)} \right) \cdot \left( {2\hat i + \hat k} \right) \\
   \Rightarrow W = 10\sqrt 3 \left( {2 + 1} \right) \\
   \Rightarrow W = 30\sqrt 3 \\ $
Therefore, the work done by the force is $ 30\sqrt 3 $ joules.

Note :
It is important to note that the change in position of the particle here only gives the displacement vector, if you take the displacement vector by taking the vector form of one of the points, then that doesn’t give the displacement vector. The force in the direction of the given vector $ \left( {\hat i + \hat j + \hat k} \right) $ can only be taken by multiplying it with the unit vector only.