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The distance from the release point to the target of a curling sheet is around 42.3 meters. A study showed that the coefficient of friction of the surface between the curling stone and ice was .30. How fast should the stone be released to get a bullseye?

The distance from the release point to the target of a curling sheet is around 42.3 meters. A study showed that the coefficient of friction of the surface between the curling stone and ice was .30. How fast should the stone be released to get a bullseye?

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The distance from the release point to the target of a curling sheet is around 42.3 meters. A study showed that the coefficient of friction of the surface between the curling stone and ice was .30. How fast should the stone be released to get a bullseye?

 

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Advanced Physics homework question answer, step 1, image 1

The Force due to friction is given by

f=-μR=-μmg

m is the mass.

From newtons second law we have

 f=ma=-μmga=-μg

hence a=mg is the deceleration that acts against the motion of the stone.

from the kinematic equation of motion, we have

v2-u2=2asv=final velocityu=initial velocitya=accelerations=displacement

given that

v=0 m/sμ=0.3g=9.8 m/s2s=42.3 m

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