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A uniform force of (3î + ĵ) newton acts on a particle of mass 2 kg. The particle is displaced from position (2î + k̂) meter to position (4î + 3ĵ − k̂) meter. The work done by the force on the particle is

1. 6 J
2. 13 J
3. 15 J
4. 9 J

Correct answer: 6 J

Solution

The work done is calculated as the dot product of the force vector and the displacement vector. The displacement vector is (4î + 3ĵ − k̂) − (2î + k̂) = (2î + 3ĵ − 2k̂). The dot product of (3î + ĵ) and (2î + 3ĵ − 2k̂) is (3×2) + (1×3) + (0×−2) = 6 + 3 + 0 = 6 J.

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