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

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