A uniform force of 3i^+j^ newton acts on a particle of mass 2 kg. Hence the particle is displaced from position 2i^+k^ meter to position 4i^+3j^−k^ meter. The work done by the force on the particle is

Here,F→=3i^+j^N Intital position, r→1=2i^+k^m Final position, r→2=4i^+3j^−k^m Diplacement, r→=4i^+3j^−k^m−2i^+k^m =2i^+3j^−2k^ m Work done, W=F→.r→=3i^+j^.2i^+3j^−2k^ =6+3=9J

A uniform force of 3i^+j^ newton acts on a particle of mass 2 kg. Hence the particle is displaced from position 2i^+k^ meter to position 4i^+3j^−k^ meter. The work done by the force on the particle is

Here,F→=3i^+j^N Intital position, r→1=2i^+k^m Final position, r→2=4i^+3j^−k^m Diplacement, r→=4i^+3j^−k^m−2i^+k^m =2i^+3j^−2k^ m Work done, W=F→.r→=3i^+j^.2i^+3j^−2k^ =6+3=9J

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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

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13J

15J

answer is C.

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Detailed Solution

$Here, \vec{F} = (3 \hat{i} + \hat{j}) N Intital position, {\vec{r}}_{1} = (2 \hat{i} + \hat{k}) m Final position, {\vec{r}}_{2} = (4 \hat{i} + 3 \hat{j} - \hat{k}) m Diplacement, \vec{r} = (4 \hat{i} + 3 \hat{j} - \hat{k}) m - (2 \hat{i} + \hat{k}) m = 2 \hat{i} + 3 \hat{j} - 2 \hat{k} m Work done, W = \vec{F} . \vec{r} = (3 \hat{i} + \hat{j}) . (2 \hat{i} + 3 \hat{j} - 2 \hat{k}) = 6 + 3 = 9 J$