Question

A book slides across a level, carpeted floor with an initial speed of 3.25 m/s and comes to rest after 3.25 m. Calculate the coefficient of kinetic friction ????k between the book and the carpet. Assume the only forces acting on the book are friction, weight, and the normal force.

Answer 1

Answer:

The value of the coefficient of kinetic friction is 0.165.

Explanation:

Given that,

Initial speed = 3.25 m/s

Distance = 3.25 m

We need to calculate the acceleration

Using third equation of motion

$v^2=u^2+2as$

$a=\dfrac{v^2-u^2}{2s}$

Where, a = acceleration

u = initial velocity

v = final velocity

s = distance

Put the value into the formula

$a=\dfrac{-(3.25)^2}{2\times3.25}$

$a=-1.625\ m/s^2$

Negative sign shows the deceleration

We need to calculate the coefficient of kinetic friction

Using frictional force

$F_{k} =k mg$....(I)

Where, k = coefficient of kinetic friction

m = mass

g = acceleration due to gravity

Using newton's second law

$F = ma$....(II)

Equating equation(I) and (II)

$k mg=ma$

$k=\dfrac{a}{g}$

$k=\dfrac{1.625}{9.8}$

$k=0.165$

Hence, The value of the coefficient of kinetic friction is 0.165.