A body is acted upon by a force F = -i-3j+4k. what is the work done by the force in displacing it from (0,0,0) to (0,0,3) m?
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Answer:
Total force =(4i^+j^−3k^)+(3i^+j^−k^)=7i^+2j^−4k^
The particle is displaced from A(i^+2j^+3k^) to B(5i^+4j^+k^)
Therefore displacement AB=(5i^+4j^+k^)−(i^+2j^+3k^)=4i^+2j^−2k^
Work done =F.AB=(7i^+2j^−4k^).(4i^+2j^−2k^)
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