InterviewSolution
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InterviewSolution
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A `3.00 kg` object has a velocity `(6.00 hati - 2.00 hatj) m//s`. What is the net work done on the object if its velocity changes to `(8.00 hati - 4.00 hatj) m//s`?A. `64.5 J`B. `64.5 J`C. `64.5 J`D. `64.5 J` |
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Answer» Correct Answer - D `vec v_(i) = (6.00 hat i - 1.00 hatj) m//s^(2)` `v_(i) = sqrt(v_(I x)^(2) + v_(I y)^(2)) = sqrt(37.0) m//s` `K_(1) = (1)/(2) mv_(1)^(2) = (1)/(2) (3.00 kg) (37.0 m^(2)//s^(2)) = 55.5 J` `vec v_(f) = 8.00 hati + 4.00 hatj` `v_(f)^(2) = vec v_(f) . vec v_(f) = 64.0 + 16.0 = 80.0 m^(2)//s^(2)` `Delta K = K_(f) - K_(i) = (1)/(2) m (v)_(f)^(2) - v_(i)^(2)` `= (3.00)/(2) (80.0) - 55.5 = 64.5 J` |
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