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I have a physics question.
Q. in the winter sport of curling, two teams alternate sliding 20 kg stones on an icy surface in an attempt to end up with the stone closet to the center of a target painted on the ice. during one turn, a player releases a stone that travels 27.9 m before coming to rest. the friction force acting on the stone is 2.0 N. what was the speed of the stone when the player released it?

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opudodennis

Answer:

2.36 m/s

Explanation:

We know that F=ma hence considering the frictional force on stone

2=20a

a=2/20= 0.1

From kinematics equations

[tex]v^{2}= u^{2}+2as[/tex]

When it comes to rest , v=0

[tex]u^{2}=2as[/tex]

[tex]u^{2}=2\times 0.1\times 27.9[/tex]

[tex]u=\sqrt {2\times 0.1\times 27.9}=2.36 m/s[/tex]

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