What is the magnitude of the horizontal force acting on the sprinter?

A 50kg sprinter, starting from rest, runs 50m in 7 s at constant acceleration.


b) what is the sprinter's average power output during the first 2.0s of his run?


c) wat is the sprinter's average power output during the final 2.0s of his run?





Can i get a detailed explanation on how to solve this please.|||A) I drew a graph for you: http://img189.imageshack.us/img189/961/s鈥?/a>





B) Power = Work done/time, Work done = Energy = force*distance, Force = mass*acceleration


With 1.0204 m/s^2 of acceleration and 50kg of mass, Force is 51.0204 Newtons


Distance travelled = acceleration*time*1/2


Energy required is 51.0204N*(1.0204*2.0seconds*1/2)=52.0612 Joules


Power = 52.0612/2=26.0306 Watts





C)Since acceleration is the same, the force exerted and therefore power by the sprinter must be the same, therefore the answer is still 26.0306 Watts.





Hope it helped your physics:D|||The acceleration a is found from s = 1/2 a t^2.


a = 2*s/ t^2 = 2*50/ 49 = 2.041 m/s^2.


Athe end of 2 second the velocity is v = at = 2.041*2 = 4.082 m/s.


Power = energy/ time (2s) = 1/2 mv^2 / t = 1/4 *50*4.082^2


=208.2 W.


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At the end of 5s its velocity is u = 2.041*5= 10.205 m/s


and at the end of 7s its velocity is v = 2.041*7= 14.287m/s


Energy gained during the last 2 s = 1/2 m [v^2 -u^2]


Power = energy/ time (2s) = 1/4 m [v^2 -u^2]


Power during the last 2 s = 0.25*50* [14.287^2 -10.205^2]


=1249.7 W.


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• 2 years ago




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