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A 3.00-kg object has a velocity 16.00 i ^ 2 2.00 j ^2 m/s.

(a) What is its kinetic energy at this moment?
(b) What is the net work done on the object if its velocity changes to 18.00 i ^ 1 4.00 j ^2 m/s? (Note: From the definition of the dot product, v 2 5 v S ? v S .)

Respuesta :

Answer:

390 J

Explanation:

m = 3 kg

u = 16 i + 2 j

(a) Magnitude of velocity = [tex]\sqrt{16^{2}+2^{2}}[/tex] = 16.1245 m/s

KEi = 1/2 m v^2 = 0.5 x 3 x 16.1245 = 390 J

(b) v = 18 i + 14 j

Magnitude of velocity =  [tex]\sqrt{18^{2}+14^{2}}[/tex] = 22.804 m/s

KEf = 1/2 m v^2 = 0.5 x 3 x 22.804 = 780 J

According to the work energy theorem

Work done = change in KE = KEf - KEi = 780 - 390 = 390 J

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