Baseball friction and distance problem-need help for future exam

Question

A 82 kg baseball player, running at 3.5 m/s, slides into first base. If the coefficient of kinetic friction is 0.40, how far does the player slide?

Answer 1

To solve this dynamics problem, let's first identify all of the forces acting on the baseball player when they are sliding into first base. The force of gravity (F_g) pulls the player towards the center of the earth while the normal force on the player by the ground (F_n) balances out the force of gravity. Additionally, the force of kinetic friction on the player by the ground (F_fr) acts in the opposite direction that the player is sliding in, slowing them down.

It is clear that the net force on the baseball player, which determines their acceleration, is dependent on the force of kinetic friction.

We can determine the force of kinetic friction by performing the following steps:

F_n = F_g = (82kg)(9.8m/s²) = 803.6N

F_fr = μ_kf(F_n) = 0.40(803.6N) = 321.4N

Then, using Newton's second law of motion, we get:

F_net = ma = 321.4N

a = 321.4N/82kg = 3.92m/s²

Now that we know the magnitude of the acceleration of the baseball player, we can solve for the total distance they travelled while sliding.

Their initial velocity is 3.5m/s, their final velocity is 0m/s, and their acceleration is -3.92m/s². The acceleration is chosen to be negative because it is directed in the opposite direction of the player's motion.

Plugging these variables into the kinematics equation V_x² = V_xo² + 2aΔx, we get the answer:

V_x² = V_xo² + 2aΔx

0m/s = (3.5m/s)² + 2(-3.92m/s²)Δx

-12.25m²/s² = (-7.84m/s²)Δx

Δx = 1.6m