Question

A baseball player rounds third base and is heading for home when he is signaled by the basecoach to slide. The player has a speed of 3.5 m/s when he begins to slide and he travels through alinear distance of 1.50 meters before coming to rest. Determine the coefficient of kinetic frictionbetween him and the ground. Show all of your work.

Answer 1

Answer:

0.4166

Explanation:

Data,

U = 0m/s

V = 3.5m/s

S = 1.50m

Fs ≤ μs N

Fs = force due to friction

μs = coefficient of friction

N = normal force

μs = Fs/N

But Fs = Mass * acceleration = Ma

N = Mass * acceleration due to gravity = Mg

a =?

From equation of motion

V² = U² + 2as

3.5² = 0² + 2 * a * 1.5

12.25 = 3a

a = 4.083 m/s²

μs = ma / mg = a/g

μs = 4.083 / 9.8

μs = 0.4166

Answer 2

Answer:

The coefficient of kinetic friction between him and the ground is 0.42.

Explanation:

given information:

the player's initial speed, $v_{0}$ = 3.5 m/s

displacement, s = 1.50 m

final speed, $v_{t}$ = 0 (coming to rest)

we know that for the linear motion,

$v_{t}$² = $v_{0}$² + 2as

where

$v_{0}$ = initial speed (m/s)

$v_{t}$ = final speed (m/s)

a = acceleration (m/s²)

s = displacement (m)

so,

$v_{t}$² = $v_{0}$² + 2as

a = $v_{t}$² - $v_{0}$²/2s

  = 0² - 3.5²/(2 x 1.5)

  = - 3.5²/3

  = - 4.083 m/s²

then, we continue to calculate the coefficient of kinetic friction by the following formula

$F = - F_{friction}$

where,

F = m a

$F_{friction}$ = μk m g

g = gravitational acceleration (m/s²)

F = force (N)

m = mass (kg)

so,

$F = - F_{friction}$

m a = - μk m g

a = - μk g

μk  = - a/g

     = - (- 4.08)/9.8

     = 0.42