Constant forces P=2^i−5^i+6^k and 4Q=−^i+2^i−^k act on a particle. Determine the work done when the particle is displaced from point A with position vector 4^i−3^i−2^k to a point B with position vector 6^i+^i−3^k
A
10
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B
−15
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C
−10
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D
15
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Solution
The correct option is A−15 The displacement vector of the particle is =→AB =2i+4j−k ...(i) Net force on the particle is =→P+4→Q =2i−5j+6k−i+2j−k =i−3j+5k...(ii) Work done =→F.→d =(2i+4j−k).(i−3j+5k) =2−12−5 =−17+2 =−15 joules.