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.

, ,

Explanation:

The equations of equilibrium for the rock are:

After some algebraic handling, the following expression is found:

Deceleration experimented by the rock is:

The distance travelled by the rock before stopping is:

And the time is:

The question is incomplete! The complete question along with answers and explanation is provided below.

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?

GIven Information:

Mass of rock = m = 20 kg

Initial speed of rock = vi = 8 m/s

Coefficient of kinetic friction = μ = 0.200

Required Information:

Average power produced due to friction = ?

Average power produced due to friction = 156.1 Watts

Explanation:

As we know the power is rate of doing work,

P = W/t

work-energy theorem relates work done and kinetic energy as

W = KE =  ½m(vf² - vi²

W = ½*20*(0 -8²)

W = -640 J

According to Newton's second law of motion,

F = ma

friction force is given by,

F = -μmg

relate the two equations

ma = -μmg

mass cancels out

a = -μg

a = -0.200*9.8

a = -1.96 m/s²

The negative sign indicates deceleration

From the kinematics equation

t = (vf - vi)/a

t = (0 - 8)/-1.96

t = -8/-1.96

t = 4.1 seconds

Therefore, the power produced to the friction is

P = 640/4.1

P = 156.1 Watts

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