Asked by Anonymous
how much negative work is done by friction on a 2000-kg car if you slow down from 20 m/s to 10 m/s?
Answers
Answered by
Scott
the work done on the car is reflected by the change in kinetic energy
w = [.5 * 2000 * 20^2] - [.5 * 2000 * 10^2]
the units are joules
w = [.5 * 2000 * 20^2] - [.5 * 2000 * 10^2]
the units are joules
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