A 90kg basketball player carries about 720J of kinetic energy when running down the court. If the coefficient of kinetic friction between good runners & the floor is .600, how far does the player skid when trying to stop? Hint: what work was done by friction in bringing the player to a stop?

Question

A 90kg basketball player carries about 720J of kinetic energy when running down the court. If the coefficient of kinetic friction  between good runners & the floor is .600, how far does the player skid when trying to stop? Hint: what work was done by friction in bringing the player to a stop?

I am so at a loss with where to go... I used Ek=1/2mV^2 to get my velocity which is 4m/s but where do I go from there?

Thanks! :)

Scott · Answer

the friction/work hint is a good one 90 * g * .6 * d = 720

Kayla · Answer

I don't think I have that formula... So mg(mu)d=Ek? making d=190.8 then I can use kinematics to get a & w=mad? :) thanks a ton :)