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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?
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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
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