Question

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 .)

Answer 1

390 J

Step-by-step explanation:

m = 3 kg

u = 16 i + 2 j

(a) Magnitude of velocity =
`\sqrt{16^(2)+2^(2)}` = 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 =
`\sqrt{18^(2)+14^(2)}` = 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