a force F =(3xi^+4j^) Newton (where x is in metres) acts on a particle which moves from a position (2m,3m) to (3m,0m) .then the work done is A) 7.5 JouleB) -12jouleC) -4.5 Joule D) +4.5 Joule
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here f=5 and s= (2-3)=-1 w=fs=5×-1=-5j so may be C is the right ans hope so
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Answer:The answer is option C -4.5J
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