Question

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`

Answer 1

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`