) A stone initially moving at 8.0 m/s on a level surface comes to rest due to friction after it travels 11 m. What is the coefficient of kinetic friction between the stone and the surface

-0.3

Explanation:

F' = μmg Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g Equation 3

From the equation of motion,

v² = u²+2as Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

answer is c

explanation: