A3.00-kg object has a velocity 1 6.00 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 1 8.00 i ^ 1 4.00 j ^

(a) The velocity of the object on the x-axis is 6 m/s, while on the y-axis is 2 m/s, so the magnitude of its velocity is the resultant of the velocities on the two axes:

And so, the kinetic energy of the object is


(b) The new velocity is 8.00 m/s on the x-axis and 4.00 m/s on the y-axis, so the magnitude of the new velocity is

And so the new kinetic energy is


So, the work done on the object is the variation of kinetic energy of the object:

Velocity=total displacement/timefrom house to store is 500 from store to neighbor is 200 therefore total displacement=500+200=700metrestherefore,700metres/910seconds=0.8metres per second

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