Question

A driver, traveling at 22.0 m/s, slows down her 2.00 x 103 kg car to stop for a red light. What work is done by the friction force against the wheels?

Answer 1

Vf^2 = 0 = Vi^2 -2ad

ad = Vi^2 / 2

F*d = m*a*d = m* Vi^2/2 = 2000*((22)^2 /2)

you can also use work/energy theorem
delta KE = (1/2)mVi^2 = Work

Answer 2

I do belive you answered my question perfectly. Thank you so much.