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(A) The given force $\vec F = (3\hat i + 4\hat j)$ is a constant force.Work done by a constant force is independent of the path taken and depends only

Vedclass · Accepted Answer

$W_1 = W_2$

Vedclass · Answer

(A) The given force $\vec F = (3\hat i + 4\hat j)$ is a constant force.Work done by a constant force is independent of the path taken and depends only on the initial and final positions.Since both paths start at $(0, 0)$ and end at $(a, a)$,the displacement vector $\vec d = (a - 0)\hat i + (a - 0)\hat j = a\hat i + a\hat j$ is the same for both paths.Therefore,the work done $W = \vec F \cdot \vec d = (3\hat i + 4\hat j) \cdot (a\hat i + a\hat j) = 3a + 4a = 7a$.Since the work done is the same for both paths,$W_1 = W_2$.

$A$ particle is moved from $(0, 0)$ to $(a, a)$ under a force $\vec F = (3\hat i + 4\hat j)$ along two paths. Path $1$ is $OP$ and path $2$ is $OQP$. Let $W_1$ and $W_2$ be the work done by this force in these two paths respectively. Then

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