# A football player, with a mass of 60.0 kg, slides on the ground after being knocked down. At the...

## Question:

A football player, with a mass of {eq}60.0 \ kg {/eq}, slides on the ground after being knocked down. At the start of the slide, the player moves at a speed of {eq}3.10 \ m / s {/eq} before coming to rest. The coefficient of friction between the player and the ground is {eq}0.700 {/eq}.

(a) Find the change in mechanical energy of the player (in {eq}J {/eq}) from the start of the slide to when he comes to rest.

(b) How far (in {eq}m {/eq}) does the player slide?

## Kinetic Energy:

When a body having some mass runs faster with some speed, then there exists energy in this body. This energy is equal to half of the product of mass of that body and velocity by which the body was moving. This energy is known as kinetic energy.

## Answer and Explanation:

Given Data

• The mass of the player is: {eq}m = 60.0\;{\rm{kg}} {/eq}.
• The speed of the player is: {eq}{v_i} = 3.10\;{\rm{m/s}} {/eq}.
• The coefficient of friction between player and ground is: {eq}\mu = 0.7 {/eq}.

(a)

The expression to calculate the change in mechanical energy is given by,

{eq}\Delta {E_m} = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 {/eq}

Here, {eq}{v_f} {/eq} is the final velocity of player which is zero as player stops at the end.

Substitute all the values in above expression.

{eq}\begin{align*} \Delta {E_m} &= \dfrac{1}{2} \times 60 \times {0^2} - \dfrac{1}{2} \times 60 \times {3.1^2}\\ & = - 288.3\;{\rm{J}} \end{align*} {/eq}

Thus the change in mechanical energy is {eq}- 288.3\;{\rm{J}} {/eq}.

(b)

The expression to calculate the distance that the player slide is given by,

{eq}\begin{align*} W &= \Delta {E_m}\\ - \mu mgd &= \Delta {E_m}\\ d &= - \dfrac{{\Delta {E_m}}}{{\mu mg}} \end{align*} {/eq}

Here {eq}W {/eq} is the work, and {eq}g {/eq} is the gravitational acceleration.

Substitute all the values in above expression.

{eq}\begin{align*} d &= - \dfrac{{\left( { - 288.3} \right)}}{{0.7 \times 60 \times 9.81}}\\ & = 0.799\;{\rm{m}} \end{align*} {/eq}

Thus the distance that the player slide is {eq}0.799\;{\rm{m}} {/eq}.