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A uniform force of $(3\hat i + \hat j)$ newton acts on a particle of mass 2 kg. Hence the particle is displaced from position $(2\hat i + \hat k)$ meter to position $(4\hat i + 3\hat j - \hat k)$ meter. The work done by the force on the particle is:
A
15 J
B
9 J
C
6 J
D
13 J
Detailed Solution
$W = \vec F\cdot\vec S = (3\hat i + \hat j)\cdot[(4 - 2)\hat i + (3 - 0)\hat j + (-1 - 1)\hat k]$
$= (3\hat i + \hat j)\cdot(2\hat i + 3\hat j - 2\hat k) = 3(2) + 1(3) + 0(-2) = 9$ J
$= (3\hat i + \hat j)\cdot(2\hat i + 3\hat j - 2\hat k) = 3(2) + 1(3) + 0(-2) = 9$ J