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Q.

Three forces (5i^+4j^+2k^)N   (−2i^−2j^+6k^)N,(i^−j−3k^)N  displaces a particle from (1m,-1m,-1m) to (1m,0,3m) and then to (3m,1m,2m). The total work done by all these forces is

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a

50J

b

25J

c

15J

d

-10J

answer is B.

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Detailed Solution

Net force  F→=F1→+F2→+F3→                 F→=4i^+j^+5k^Displacement Δr→=r3→−r→1=3i^+j^+2k^−(i^−j^−k^)Δr→=2i^+2j^+3k^W=F→⋅Δr→=(4i^+j^+5k^)⋅(2i^+2j^+3k^)=8+2+15=25J
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