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Firlakuza [10]
2 years ago
9

A 3.00-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 w

ork done on the object if its velocity changes to 1 8.00 i ^ 1 4.00 j ^
Physics
1 answer:
tatyana61 [14]2 years ago
7 0
(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:
v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s
And so, the kinetic energy of the object is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J

(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
v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s
And so the new kinetic energy is
K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J

So, the work done on the object is the variation of kinetic energy of the object:
W=\Delta K=120 J-60 J=60 J
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