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A particle of mass `0.5kg` travels in a straight line with velocity `v=ax^(3//2)` where `a=5m^(-1//2)s^-1`. What is the work done by the net force during its displacement from `x=0` to `x=2m`?A. 50 JB. 20 JC. 80 JD. `45.5 J` |
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Answer» Correct Answer - A `W = Delta KE = (1)/(2) mv^(2) - (1)/(2) m u^(2)`. |
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