There is a scene that constantly shows up in movies about the Old West. A cowboy

Question

There is a scene that constantly shows up in movies about the Old West. A cowboy sits down at a bar and orders a drink. The bartender, at the other end of the bar, fills a glass and slides it down the bar with it stopping right in front of the thirsty patron. If the filled mug has a mass of 0.75 kg and the coefficient of kinetic friction between the bar top and the mug is 0.35, with what speed must the bartender release the mug for it to slide 2.5 meters and come to rest in front of the patron? Also, with what speed must the patron push the empty mug (mass of 0.25 kg) to return it to rest in front of the bartender? Be sure to specify what condition(s) allow you to use the method that you have chosen.

Explanation / Answer

friction force, f = mu*mg = 0.35*0.75*9.8 = 2.5725N
initial v= ?
final v = 0
deceleration = f/m = 3.43 m/s/s
s = 2.5m
2*a*s = v^2
v = sqroot(2*3.43*2.5) = 4.1412 m/s

for the other way round, v = sqroot(2as) = sqroot(2*mu*g*s) = sqroot(2*0.35*9.8*2.5) = 4.1412 m/s

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