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s. How much work is done on the opponent?

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asked Mar 18, 2019 188k views

5 votes

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?

Physics
college

Robaticus asked

by Robaticus

8.4k points

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1 Answer

4 votes

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

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

Mohammed Nagoor answered Mar 24, 2019