Answer :
Final answer:
The initial kinetic energy of the 3.00-kg object traveling at a velocity of 2.00 m/s is 6.00 Joules. When the object's velocity changed to 4.47 m/s, its kinetic energy became 30.02 Joules. Hence, the net work done on the object is 24.02 Joules.
Explanation:
The kinetic energy of any object can be calculated using the formula KE = 0.5 * m * v^2, where m is the object's mass and v is its velocity. For the 3.00-kg object with a velocity of 6.00 i ^ 2 and 2.00 j ^2 m/s, its velocity magnitude would be the square root of (6.00^2 + 2.00^2), which is 2.00 m/s. Plugging the values into the formula, the kinetic energy (a) would be 0.5 * 3.00 * 2.00^2 = 6.00 Joules.
The net work done on an object (b) can be obtained by finding the change in kinetic energy when the object’s velocity changes to 8.00 I and 4.00 j. The final velocity's magnitude would be the square root of (8.00^2 + 4.00^2), which is 4.47 m/s. Hence, the final kinetic energy is 0.5 * 3.00 * 4.47^2 = 30.02 Joules. Therefore, the net work done equals the change in kinetic energy, which is 30.02 - 6.00 = 24.02 Joules.
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(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:
[tex]v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s [/tex]
And so, the kinetic energy of the object is
[tex]K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J [/tex]
(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
[tex]v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s [/tex]
And so the new kinetic energy is
[tex]K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J [/tex]
So, the work done on the object is the variation of kinetic energy of the object:
[tex]W=\Delta K=120 J-60 J=60 J[/tex]
[tex]v= \sqrt{(6.00m/s)^2+(2.00 m/s)^2}=6.32 m/s [/tex]
And so, the kinetic energy of the object is
[tex]K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(6.32 m/s)^2=60 J [/tex]
(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
[tex]v= \sqrt{(8.00 m/s)^2+(4.00 m/s)^2}=8.94 m/s [/tex]
And so the new kinetic energy is
[tex]K= \frac{1}{2}mv^2= \frac{1}{2}(3.00 kg)(8.94 m/s)^2=120 J [/tex]
So, the work done on the object is the variation of kinetic energy of the object:
[tex]W=\Delta K=120 J-60 J=60 J[/tex]
