Question

A baseball player slides into home base and experiences coefficients of friction of μ8 = 0.63 and μk = 0.49. The normal force acting on the player is 660N. How much frictional force acts on the player?Round your answer to the nearest hundredth if necessary. Ff = ________ N

Answer 1

Answer:

323.4 N

Explanation:

Given that,

The coefficient of static friction, $\mu_s=0.63$

The coefficient of kinetic friction, $\mu_k=0.49$

We need to find the frictional force acting on the player. As the baseball player is sliding into home base, we will use the coefficient of kinetic friction to find the frictional force.

$F=\mu _kN\\\\=0.49\times 660\\\\=323.4\ N$

So, the frictional force acting on the player is 323.4 N.