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An 80 kg skateboarder moving at 3 m/s pushes off with her back foot to move faster. If her velocity increases to 5 m/s, what is her change in kinetic energy as a result? J How much work did she perform?

Respuesta :

skyluke89
1) The kinetic energy of an object is given by:
[tex]K= \frac{1}{2}mv^2 [/tex]
where m is the object's mass and v its speed.

By using this equation, we find the initial kinetic energy of the skateboarder:
[tex]K_i= \frac{1}{2}(80 kg)(3 m/s)^2=360 J [/tex]
and the final kinetic energy as well:
[tex]K_f= \frac{1}{2}(80 kg)(5 m/s)^2=1000 J [/tex]

So, her change in kinetic energy is
[tex]\Delta K=K_f-K_i=1000 J-360 J=640 J[/tex]

2) The work-energy theorem states that the work done to increase the speed of an object is equal to the variation of kinetic energy of the object:
[tex]W=\Delta K[/tex]
Therefore, the work done by the skateboarder is
[tex]W=\Delta K=640 J[/tex]
raphealnwobi

The skateboarder  change in kinetic energy is 160 J and the work done is 160 J

The initial velocity (u) = 3 m/s, final velocity (v) = 5 m/s, mass (m) = 80 kg.

The change in Kinetic energy = (1/2) * m * (v - u)² = 0.5 * 80 * (5 - 3)² = 160 J

The work done = change in kinetic energy = 160 J

Hence the skateboarder  change in kinetic energy is 160 J and the work done is 160 J

Find out more at: https://brainly.com/question/19210991

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