A particle moves from position r1¯=3i^+2j^−6k^ to position r→2=14i^+13j^+9k^ under the action of force F→=4 i⏞+ j⏞+3k⏞. The work done will be

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100 J

50 J

200 J

75 J

answer is A.

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Detailed Solution

W=F⇀⋅r2⇀-r1⇀=(4i^+j^+3k^)(11i^+11j^+15k^)W=44+11+45=100 Joule

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