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A 20.0-kg rock is sliding on a rough, horizontal surface at 8.00 m/s and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is 0.200. What average power is produced by friction as the rock stops?

Answer :

Answer:

Therefore,

The average power produced by friction as the rock stops is 156.86 Watts.

Explanation:

Given:

mass = m =20.0 kg

initial velocity,

[tex]v_{i}=8.00\ m/s[/tex]

coefficient of kinetic friction,

[tex]\mu_{k}=0.200\ m/s[/tex]

To Find:

average power, P =?

Solution:

According to Newton's Second Law,

[tex]Force=mass\times Acceleration[/tex]

Also Friction force,

[tex]Friction\ Force=-\mu_{k}mg[/tex]

Therefore,

[tex]Force=Friction\ Force[/tex]

Substituting the values we get

[tex]m\times a=-\mu_{k}mg\\\\a=-0.2\times 9.8=-1.96\ m/s^{2}[/tex]

Since Rock is getting stopped therefore acceleration is negative.

Final velocity will also be ZERO.

[tex]v_{f}=0\ m/s[/tex]

Now by Kinematic Equation

[tex]v_{f}=v_{i}+at[/tex]

Substituting the values we will get time 't',

[tex]t=-\dfrac{8}{-1.96}=4.08\ s[/tex]

Now  according to work theorem ,

[tex]W=Change\ in\ Kinetic\ Energy[/tex]

which is equal to

[tex]W=\dfrac{1}{2}m((v_{f})^{2}-(v_{i})^{2})[/tex]

Substituting the values we get

[tex]W=-\dfrac{1}{2}\times 20\times 8^{2}=-640\ Joules[/tex]

Now ,

[tex]Power=|\dfrac{Work}{time}|[/tex]

Substituting the values we get

[tex]Power=|\dfrac{-640}{4.08}|=156.86\ Watt[/tex]

Therefore,

The average power produced by friction as the rock stops is 156.86 Watts.

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