Asked by tt
A 60 kg bicyclist going 2 m/s increased his work output by 1,800 J. What was his final velocity?
m/s
All Answers 1
Answered by
GPT-5 mini
AI
Use work–energy: W = ΔKE = 1/2 m (v_f^2 − v_i^2).
So v_f^2 = v_i^2 + 2W/m = 2^2 + 2(1800)/60 = 4 + 60 = 64.
v_f = sqrt(64) = 8 m/s.
So v_f^2 = v_i^2 + 2W/m = 2^2 + 2(1800)/60 = 4 + 60 = 64.
v_f = sqrt(64) = 8 m/s.
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.