JEE MAIN - Physics (2004 - No. 43)
A force $$\overrightarrow F = \left( {5\overrightarrow i + 3\overrightarrow j + 2\overrightarrow k } \right)N$$ is applied over a particle which displaces it from its origin to the point $$\overrightarrow r = \left( {2\overrightarrow i - \overrightarrow j } \right)m.$$ The work done on the particle in joules is
$$+10$$
$$+7$$
$$-7$$
$$+13$$
Explanation
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:
$$W = \overrightarrow F \cdot \overrightarrow r$$
We can substitute the given vectors into this expression:
$$W = \overrightarrow F .\overrightarrow r $$
$$= \left( {5\widehat i + 3\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j} \right)$$
$$=10-3=7$$ J
We can substitute the given vectors into this expression:
$$W = \overrightarrow F .\overrightarrow r $$
$$= \left( {5\widehat i + 3\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j} \right)$$
$$=10-3=7$$ J
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