# A baseball player rounds second base in an arc with a radius of curvature of 6.52m at a speed of 6.8m/s. Please Help!?

If he weights 1039 N what is the centripal force and the coefficient of kinetic friction between the shoes of the player and the ground? I figured out that the centripetal force is 752N but I don t know the friction.

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Since the baseball player is moving at a constant speed, the friction force is equal to the centripetal force. To solve this problem, I will need to determine the exact number for the centripetal force. This is good way to check your answer.

Mass = 1039 ÷ 9.8

This is approximately 106 kg.

Fc = (1039 ÷ 9.8) * 6.8^2 ÷ 6.52 = 48,043.36 ÷ 63.896

This is approximately 752 N.

Ff = μ * 1039

μ * 1039 = 48,043.36 ÷ 63.896

μ = 48,043.36 ÷ 66,387.994

The coefficient of friction is approximately 0.72. I hope this is helpful for you.

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• W*μ = W/g*V^2/r

μ*g*r = V^2

μ = V^2/(g*r) = 6.8^2/(6.52*9.806) = 0.723

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• The centripetal force that you calculated is the force required to follow that arc at that speed. The only possible supplier of that centripetal force is the friction with the ground. So the friction provided those 752 N.

However, that information would be able to give you the coefficient of kinetic friction only if that speed was the fastest speed that would allow the runner to follow that arc. But the question did not tell us that. With the facts that we have, it is possible that a faster runner (with the same weight) could follow that arc at a higher speed. If we assume that was the limiting speed, then

mu_k = 752 N/1039 N

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