100.0-kg football player is running at 6.0 m/s He is tackled and stops in 1.3 seconds_ a) What was the original kinetic energy ofthe player? b) What was the average power required to stop him?
Added by Dana L.
Step 1
0 kg Velocity (v) = 6.0 m/s Plugging in the values: \( KE = \frac{1}{2} \times 100.0 \times (6.0)^2 \) \( KE = \frac{1}{2} \times 100.0 \times 36 \) \( KE = 1800 \, J \) Show more…
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