JEE MAIN - Mathematics (2004 - No. 25)
A particle acted on by constant forces $$4\widehat i + \widehat j - 3\widehat k$$ and $$3\widehat i + \widehat j - \widehat k$$ is displaced from the point $$\widehat i + 2\widehat j + 3\widehat k$$ to the point $$\,5\widehat i + 4\widehat j + \widehat k.$$ The total work done by the forces is :
$$50$$ units
$$20$$ units
$$30$$ units
$$40$$ units
Erläuterung
The work done by a force on a particle is given by the dot product of the force and the displacement vector of the particle. The displacement vector can be found by subtracting the initial position from the final position:
$$\mathbf{displacement} = \mathbf{final\ position} - \mathbf{initial\ position} = (5\widehat i + 4\widehat j + \widehat k) - (\widehat i + 2\widehat j + 3\widehat k) = 4\widehat i + 2\widehat j - 2\widehat k$$
The total work done by the two forces is equal to the sum of the work done by each force. The work done by each force can be calculated as the dot product of the force and the displacement:
$$\mathbf{work\ done\ by\ force\ 1} = (4\widehat i + \widehat j - 3\widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k) = 4 \cdot 4 + 1 \cdot 2 - 3 \cdot -2 = 16 + 2 + 6 = 24$$
$$\mathbf{work\ done\ by\ force\ 2} = (3\widehat i + \widehat j - \widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k) = 3 \cdot 4 + 1 \cdot 2 - 1 \cdot -2 = 12 + 2 + 2 = 16$$
The total work done by the forces is the sum of the work done by each force:
$$\mathbf{total\ work\ done} = 24 + 16 = 40$$
Therefore, the total work done by the forces is 40 J (joules) or 40 units.
$$\mathbf{displacement} = \mathbf{final\ position} - \mathbf{initial\ position} = (5\widehat i + 4\widehat j + \widehat k) - (\widehat i + 2\widehat j + 3\widehat k) = 4\widehat i + 2\widehat j - 2\widehat k$$
The total work done by the two forces is equal to the sum of the work done by each force. The work done by each force can be calculated as the dot product of the force and the displacement:
$$\mathbf{work\ done\ by\ force\ 1} = (4\widehat i + \widehat j - 3\widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k) = 4 \cdot 4 + 1 \cdot 2 - 3 \cdot -2 = 16 + 2 + 6 = 24$$
$$\mathbf{work\ done\ by\ force\ 2} = (3\widehat i + \widehat j - \widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k) = 3 \cdot 4 + 1 \cdot 2 - 1 \cdot -2 = 12 + 2 + 2 = 16$$
The total work done by the forces is the sum of the work done by each force:
$$\mathbf{total\ work\ done} = 24 + 16 = 40$$
Therefore, the total work done by the forces is 40 J (joules) or 40 units.