KE=W(fr)
mv₀²/2=F(fr)s=μmgs
s= v₀²/2μg
v=v₀-at
v=0 => a= v₀/t
s=v₀t-at²/2=
= v₀t- v₀t²/2t= v₀t/2
t=2s/v₀=2v₀²/2μgv₀=
=v₀/μg=6/0.1•9.8 =6.12 s
A puck of mass .30 kg is sliding along an essentially frictionless patch of ice at 6.0 m/s. It encounters a rough ice patch with a coefficient of kinetic friction of .10. How long will it take for the puck to stop?
1 answer
1 answer