A 103-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk = 0.555. (a) What is the magnitude of the frictional force? (b) If the player comes to rest after 1.05 s, what is his initial speed?
See previous post: Sat, 8-15-15, 7:29 AM
To solve this problem, we need to use the concepts of frictional force and kinetic friction and apply Newton's second law of motion.
(a) To find the magnitude of the frictional force, we can use the formula for kinetic friction:
frictional force = coefficient of kinetic friction * normal force
The normal force is equal to the weight of the baseball player, which can be calculated using the formula:
weight = mass * gravitational acceleration
Since the gravitational acceleration is approximately 9.8 m/s² on Earth, we can calculate the weight of the player:
weight = 103 kg * 9.8 m/s² = 1009.4 N
Now, we can calculate the frictional force:
frictional force = 0.555 * 1009.4 N ≈ 560.1 N
Therefore, the magnitude of the frictional force is approximately 560.1 N.
(b) In order to find the player's initial speed, we can use the equation of motion:
final velocity = initial velocity + (acceleration * time)
In this case, the final velocity is 0 m/s (since the player comes to rest), the acceleration is the acceleration due to friction, and the time is given as 1.05 s.
Using Newton's second law, we know that the acceleration is equal to the net force divided by the mass:
acceleration = net force / mass
Since the net force acting on the player is the frictional force and the mass is 103 kg, we can substitute these values into the equation:
acceleration = 560.1 N / 103 kg ≈ 5.43 m/s²
Now, we can rearrange the equation of motion to solve for the initial velocity:
0 m/s = initial velocity + (5.43 m/s² * 1.05 s)
Simplifying the equation:
initial velocity = -5.43 m/s² * 1.05 s ≈ -5.7 m/s
Therefore, the player's initial speed is approximately 5.7 m/s in the opposite direction of motion.