Question

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?

Answer 1

640 & 640 are correct

Answer 2

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

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

So, her change in kinetic energy is
$\Delta K=K_f-K_i=1000 J-360 J=640 J$

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:
$W=\Delta K$
Therefore, the work done by the skateboarder is
$W=\Delta K=640 J$