Question

a football player slows an 85 kg opponent down from 6 m/s to 2 m/s. How much work is done on the opponent?

Answer 1

Work is defined to be equal to change of kinetic energy:
W=ΔEk


`W = E_(kinetic-before) -E_(kinetic-after) \\ W = (1)/(2) *m* v_(1) ^(2) -(1)/(2) *m* v_(2) ^(2) \\ W=(1)/(2) *m* (v_(1) ^(2) - v_(2) ^(2) ) \\ W=(1)/(2) * 85 * ( 6^(2) - 2^(2) ) \\ W=1360J`