Question

A uniform force of $ \left( {3\hat i + \hat j} \right) $ newton acts on a particle of mass 2 kg. Hence the particle is displaced from position $ \left( {2\hat i + \hat k} \right) $ meter to position $ \left( {4\hat i + 3\hat j - \hat k} \right) $ meter. The work done by the force on the particle is
(A) 9 J
(B) 6 J
(C) 13 J
(D) 15 J

Answer 1

Hint
The work done is given by the dot product of the force and the displacement. So in the question the force is given and we need to find the displacement which is the difference between the initial and the final position.
In this solution we will be using the following formula,
 $\Rightarrow W = \vec F \cdot \vec S $
Where $ \vec F $ is the force on the particle
 $ \vec S $ is the displacement of the particle from the initial position and
 $ W $ is the work done by the force on the particle.

Complete step by step answer
The work done by a force which is acting on a particle is given by the dot product of the force and the displacement that is caused by the particle due to this force.
Here we are provided with the initial and the final position of the particle. So we can find the displacement of the particle which is given by the initial position subtracted from the final position.
Therefore according to the question, the initial position is $ \left( {2\hat i + \hat k} \right)m $ and the final position is $ \left( {4\hat i + 3\hat j - \hat k} \right)m $ . Hence, the displacement is,
 $\Rightarrow \vec S = \left( {4\hat i + 3\hat j - \hat k} \right) - \left( {2\hat i + \hat k} \right) $
On doing the subtraction we get,
 $\Rightarrow \vec S = \left( {4 - 2} \right)\hat i + 3\hat j + \left( { - 1 - 1} \right)\hat k $
Hence we get the displacement as,
 $\Rightarrow \vec S = \left( {2\hat i + 3\hat j - 2\hat k} \right)m $
The force given in the question is $ \vec F = \left( {3\hat i + \hat j} \right)N $
Therefore the work done will be the dot product as,
 $\Rightarrow W = \vec F \cdot \vec S $
Substituting the values of the force and the displacement we get,
 $\Rightarrow W = \left( {3\hat i + \hat j} \right) \cdot \left( {2\hat i + 3\hat j - 2\hat k} \right) $
On doing the dot product we have,
 $\Rightarrow W = \left( {3 \times 2} \right) + \left( {3 \times 1} \right) + (0 \times - 2) $
Therefore, on calculating we get
 $\Rightarrow W = 6 + 3 = 9J $
Therefore the work done will be $ 9J $ .
So the correct option will be (A).

Note
When the force is constant and the angle between the force and the displacement is given by $ \theta $ then the work done is given by the formula,
 $\Rightarrow W = \left| {\vec F} \right|\left| {\vec S} \right|\cos \theta $ where $ \left| {\vec F} \right| $ and $ \left| {\vec S} \right| $ are the magnitudes of the force and displacement vectors.