Morgan asked in Science & MathematicsPhysics · 1 year ago

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.

Relevance
• 1 year ago

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.

• oubaas
Lv 7
1 year ago

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