Question #253149

An athlete reaches the finish line at a velocity of 33 m s−1. She then applies a minimum braking force of 240 N as she moves along the uphill section of track to help her come to a stop.

Calculate the minimum uphill length of track L that should be available for braking. You should ignore all frictional forces other than those applied by the athlete.

mass of sledge and athlete = 95 kg


Expert's answer

ma=−Fbr=−240ma=-F_{br}=-240 N

mv=−Fbrt+cmv=-F_{br}t+c


for t=0:

v0=33 m/s  ⟹  c=v0mv_0=33\ m/s \implies c=v_0m


mL=−Fbrt2/2+v0mtmL=-F_{br}t^2/2+v_0mt


L(t)=−Fbrt2/(2m)+v0tL(t)=-F_{br}t^2/(2m)+v_0t


L′(t)=Fbrt/m+v0=0L'(t)=F_{br}t/m+v_0=0


t=mv0/Fbr=95⋅33/240=13t=mv_0/F_{br}=95\cdot 33/240=13 s


Lmin=33⋅13−240⋅132/(2⋅95)=215.5L_{min}=33\cdot 13-240\cdot 13^2/(2\cdot 95)=215.5 m


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