Question

An 75 Kg soccer player is running at 3.0 m/s and slide tackles an opponent. He slides a distance of 4.0 m before coming to rest. a. Calculate the soccer players’ deceleration (negative acceleration). b.Calculate the force acting on the player slowing him down. c.The force slowing the player down comes from kinetic friction. Calculate the coefficient of kinetic friction.

Answer 1

Given data:

* The mass of the soccer player is 75 kg.

* The velocity of the soccer player is 3 m/s.

* The final velocity of the soccer player is 0 m/s.

* The distance traveled before coming to the rest is 4 m.

Solution:

(a). By the kinematics equation, the deceleration of the soccer player in terms of the velocity and distance of the soccer player is,

`v^2-u^2=2aS`

where v is the final velocity, u is the initial velocity, a is the deceleration, and S is the distance tarveled before coming to the rest,

Substituting the known values,

`\begin{gathered} 0-3^2=2a*4 \\ -9=8a \\ a=(-9)/(8) \\ a=-1.125ms^(-2) \end{gathered}`

Thus, the deceleration of the soccer player is 1.125 meter per second squared.

(b). By the Newton's second law, the force acting on the soccer player in terms of the deceleration is,

`F=ma`

where m is the mass of the soccer player, and F is the force acting,

Substituting the known values,

`\begin{gathered} F=75*(-1.125) \\ F=-84.38\text{ N} \end{gathered}`

Here, the negative sign indicates the force is acting opposite to the direction of motion of hockey player,

Thus, the force acting on the soccer player is -84.38 N.

(c). The kinetic friction force acting on the soccer player to stop is,

`F=-\mu mg`

where,

`\begin{gathered} \mu\text{ is the coefficient of kinetic friction, and g is the acceleration due to} \\ \text{gravity} \end{gathered}`

Substituting the known values,

`\begin{gathered} -84.38=-\mu*75*9.8 \\ 84.38=\mu*735 \\ \mu=(84.38)/(735) \\ \mu=0.115 \end{gathered}`

Thus, the coefficient of kinetic friction is 0.115.