A Baseball player rounds third base and is heading for home when he is signalled by the base coach to slide. The player has a speed of 3.5 m/s when he begins to slide and he travels through a linear distance of 1.50 m before coming to rest. Determine the coefficient of kinetic friction between him and the ground.

Respuesta :

Answer:

0.42

Explanation:

The force of friction exerted on the player is:

[tex]F_f = -\mu mg[/tex] (1)

where

[tex]\mu[/tex] is the coefficient of kinetic friction

m is the mass of the player

g is the acceleration due to gravity

According to Newton's second law,

[tex]F_f = ma[/tex] (2)

where a is the player's acceleration. Substituting (1) into (2),

[tex]a=-\mu g[/tex] (3)

So in order to find [tex]\mu[/tex], we need to find the acceleration. We know the following:

u = 3.5 m/s is the initial speed of the player

v = 0 is the final speed

d = 1.50 m is the distance travelled

So we can use the following equation

[tex]v^2-u^2 = 2ad[/tex]

and solving for a

[tex]a=\frac{v^2-u^2}{2d}=\frac{0-(3.5 m/s)^2}{2(1.50 m)}=-4.1 m/s^2[/tex]

Substituting this into (3), we find the coefficient of kinetic friction:

[tex]\mu = -\frac{a}{g}=-\frac{-4.1 m/s^2}{9.8 m/s^2}=0.42[/tex]

A force that acts among sliding parts is referred to as kinetic friction force. The coefficient of kinetic friction between him and the ground is 0.415.

What is kinetic friction?

A force that acts among sliding parts is referred to as kinetic friction. A body moving on the surface is subjected to a force that opposes its progressive motion.

The size of the force will be determined by the kinetic friction coefficient between the two materials.

The given data in the problem is;

μ is the coefficient of kinetic friction=?

m is the mass of the player

g is the acceleration due to gravity= 9.81 m/s²

v is the speed of the player=3.5m/sec

x is the linear distance travelled=1.50 m

According to Newton's third equation of motion;

[tex]\rm v^2=u^2+2ax\\\\ \rm v^2-u^2=2ax\\\\ \rm a=\frac{ v^2-u^2}{2x} \\\\ \rm a=\frac{ 0^2-(3.5)^2}{2\times1.50}\\\\ \rm a=-4.08m/s^2[/tex]

The formula for the acceleration for the kinetic friction will be;

[tex]\rm a=\mu g \\\\ \rm \mu=\frac{a}{g} \\\\ \rm \mu=\frac{-4.08}{9*.81}\\\\ \rm \mu=0.415[/tex]

Hence the coefficient of kinetic friction between him and the ground is 0.415.

To learn more about the coefficient of kinetic friction refer to;

https://brainly.com/question/19180015