The game of curling involves sliding a heavy stone (to which a handle is attache

Question

The game of curling involves sliding a heavy stone (to which a handle is attached) on ice so that it stops at a certain point. For these problems, use the local value of g, 9.809 m/s^2.

1.) If the curling stone was initially sliding at 3.03 m/s and stopped precisely 11.5 meters from its launch point, then the coefficient of friction (the ratio between the magnitude of the frictional force and the magnitude of the normal force) was:

2.) Curlers commonly run in front of their stones, brushing the ice in front of the moving stones with a broom. If a stone were again released with an initial speed of 3.03 m/s and, preceded by a feverishly sweeping teammate, now goes 1.35 meters further, we would say the sweeper had reduced the coefficient of friction to:

3.) Two curling contestants dispute possession of a stone of mass 24.5 kg. The mass of contestant A is 64.7 kg, whereas that of contestant B is 117.4 kg. Contestant A pulls on the stone with a horizontal force of 15.1 N, but is unable to break contestant B

Explanation / Answer

1) F = u m g
a = ug

v^2 = v0^2 + 2 a x

0 = 3.03^2 -2*u*9.81*11.5

u=0.0407

2)
0 = 3.03^2 -2*u*9.81*(11.5+1.35)

u=0.0364

3)
a) have to cancel so 15.1 N

b) 15.1 = u m g

u = 15.1/(m g) = 15.1/(64.7*9.81)=0.0238

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