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A force of `(3hati-1.5hatj)N` acts on `5kg` body. The body is at a position of `(2hati-3hatj)m` and is travelling at `4m//s`. The force acts on the body until it is at the position `(hati+5hatj)m`. Calculate final.speed.A. `sqrt(20)m//s`B. `sqrt(10)m//s`C. `sqrt(40)m//s`D. `sqrt(50)m//s` |
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Answer» Correct Answer - B A force of ………… `5 kg` `vecF = 3hati-1.5hatj` `veca=(3)/(5)hati-(3)/(10)hatj` `vecs=(hati+5hatj)-(2hati-3hatj)=(-hati+8hatj)m` `w = vecF.vecs= (1)/(2)m(v^(2)-u^(2))` `v = sqrt(10)m//s` |
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