## 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 fricti

Question

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?

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1 year 2021-08-31T21:50:00+00:00 1 Answers 94 views 0

Therefore,

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

Explanation:

Given:

mass = m =20.0 kg

initial velocity,

$$v_{i}=8.00\ m/s$$

coefficient of kinetic friction,

$$\mu_{k}=0.200\ m/s$$

To Find:

average power, P =?

Solution:

According to Newton’s Second Law,

$$Force=mass\times Acceleration$$

Also Friction force,

$$Friction\ Force=-\mu_{k}mg$$

Therefore,

$$Force=Friction\ Force$$

Substituting the values we get

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

Since Rock is getting stopped therefore acceleration is negative.

Final velocity will also be ZERO.

$$v_{f}=0\ m/s$$

Now by Kinematic Equation

$$v_{f}=v_{i}+at$$

Substituting the values we will get time ‘t’,

$$t=-\dfrac{8}{-1.96}=4.08\ s$$

Now  according to work theorem ,

$$W=Change\ in\ Kinetic\ Energy$$

which is equal to

$$W=\dfrac{1}{2}m((v_{f})^{2}-(v_{i})^{2})$$

Substituting the values we get

$$W=-\dfrac{1}{2}\times 20\times 8^{2}=-640\ Joules$$

Now ,

$$Power=|\dfrac{Work}{time}|$$

Substituting the values we get

$$Power=|\dfrac{-640}{4.08}|=156.86\ Watt$$

Therefore,

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