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

1) The kinetic energy of an object is given by:

`K= (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= (1)/(2)(80 kg)(3 m/s)^2=360 J`
and the final kinetic energy as well:

`K_f= (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`