Oracle Posted February 25, 2010 Posted February 25, 2010 I'm totally new to learning physics and was presented with this problem but the answer was never recorded during the lecture so I'm just checking on here if I solved it correctly or what went wrong.. "On the Monark Bike - What is the force required to determine a power output of 100W when pedaling at 50rpm?" 20N = 4kg x 5m/sec 100W = 6000J / 60sec *just to note 1 revolution of the pedals will move the bike 6m
Mr Skeptic Posted February 27, 2010 Posted February 27, 2010 Are we assuming the bike is moving at a constant speed doing 100W worth of friction to cancel the power input? Otherwise, your force may have to vary depending on time. I think your answer will have to be a torque, unless you know the size of the pedals.
Iggy Posted February 27, 2010 Posted February 27, 2010 Your answer of 20 N is correct. F = Pt/d F is force in Newtons, P is power in watts (100), t is time (1 sec), and d is distance (5 m). F = 100*1/5 = 20 Newtons.
swansont Posted February 27, 2010 Posted February 27, 2010 20N = 4kg x 5m/sec This part is wrong — your units are incorrect. kg-m/sec does not give you Newtons.
Oracle Posted February 28, 2010 Author Posted February 28, 2010 (edited) Thanks for the feedback. This part is wrong — your units are incorrect. kg-m/sec does not give you Newtons. How should that part be written? in my notes it only has F=m.a2 (the 2 is a negative power - I don't understand that part) Merged post follows: Consecutive posts mergedAre we assuming the bike is moving at a constant speed doing 100W worth of friction to cancel the power input? Otherwise, your force may have to vary depending on time. I think your answer will have to be a torque, unless you know the size of the pedals. Yeah not that much detail, just a constant speed. Edited February 28, 2010 by Oracle Consecutive posts merged.
swansont Posted February 28, 2010 Posted February 28, 2010 F = ma a is acceleration (in units of m/s^2), which is the rate of change of velocity
Mr Skeptic Posted February 28, 2010 Posted February 28, 2010 Yeah, the -2 power would be for the seconds.
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