A baseball player stealing second base runs at 7.2 m/s . If he slides the last 3.8 m , slowing to a stop at the base, what's the coefficient of kinetic friction between player and ground?
initial speed u=7.2m/s
final speed v=0
distance traveled s=3.8m
coefficient of kinetic friction muK =?
We know that v^2-U^2=2as
0^2-U^2=2*(-muk^g)s
muk=(U^2)/(2*g*s)
= (7.2*7.2)/(2*9.8*3.8)
=0.696(no units)
= 0.70 (n0 units and two significant figures)
To find the coefficient of kinetic friction between the player and the ground, we can use the equation:
frictional force = coefficient of kinetic friction × normal force
First, let's find the initial speed of the player when he reaches the base. Since the player slows down from 7.2 m/s to 0 m/s over a distance of 3.8 m, we can use the equation:
vf^2 = vi^2 + 2aΔx
where:
vf = final velocity (0 m/s since the player comes to a stop)
vi = initial velocity (7.2 m/s)
a = acceleration (unknown)
Δx = distance (3.8 m)
Rearranging the equation, we have:
0 = (7.2 m/s)^2 + 2a(3.8 m)
Simplifying the equation:
0 = 51.84 m^2/s^2 + 7.6a
Now, let's solve for the acceleration (a):
7.6a = -51.84 m^2/s^2
a = -51.84 m^2/s^2 ÷ 7.6
a ≈ -6.816 m/s^2
Since the acceleration is negative, it means the player is slowing down. The negative sign indicates the direction of the acceleration, opposite to the initial velocity.
Now, we can calculate the normal force (N) acting on the player. The normal force is equal to the player's weight (mass × gravitational acceleration) when there are no other vertical forces acting on him.
Next, we need the mass of the player. Let's assume it is 80 kg.
Weight (W) = mass × gravitational acceleration
W = 80 kg × 9.8 m/s^2
W = 784 N
Since the normal force (N) is equal to the weight (W), N = 784 N.
Finally, substituting the values of acceleration (a = -6.816 m/s^2) and normal force (N = 784 N) into the equation:
frictional force = coefficient of kinetic friction × normal force
μk × 784 N = -6.816 m/s^2 × 80 kg
Simplifying the equation:
μk = (-6.816 m/s^2 × 80 kg) ÷ 784 N
μk ≈ -0.699
The coefficient of kinetic friction between the player and the ground is approximately 0.699 (rounded to three significant figures).
Note: The negative sign in the coefficient of kinetic friction (-0.699) indicates that the force of friction opposes the motion of the player. We ignore the negative sign when dealing with the magnitude of the coefficient.