Question

When you push a 1.70 kg book resting on a tabletop it takes 2.60 N to start the book sliding. Once it is sliding, however, it takes only 1.50 N to keep the book moving with constant speed. What are the coefficients of static and kinetic friction between the book and the tabletop

Answer 1

Answer:

Coefficient of static friction  ,μs = 0.15

Coefficient of kinetic friction  ,μk = 0.089

Explanation:

Given that

m = 1.7 kg

Static friction :

The force required to start the motion F= 2.6 N

We know that when book is in rest condition then static friction force act on it.

We know that the maximum value of the static friction force

fr = μ m g

At limiting condition

F= fr

2.6 =  μ m g

2.6 =  μ x 1.7 x 9.81

$\mu=\dfrac{2.6}{1.7\times 9.81}$

μ = 0.15

Kinetic friction :

F= 1.5 N

When the book is in moving condition then kinetic friction force act on it.

We know that the maximum value of the kinetic friction force

fr = μ m g

F= fr

=  μ m g

1.5 =  μ x 1.7 x 9.81

$\mu=\dfrac{1.5}{1.7\times 9.81}$

μ = 0.089

Coefficient of static friction  ,μs = 0.15

Coefficient of kinetic friction  ,μk = 0.089

Answer 2

Answer:

Explanation:

mass of book, m = 1.7 kg

Force to start the book, F = 2.6 N

Force to slide the book, f' = 1.5 N

Let μs and μk is the coefficient of static friction and kinetic friction, respectively.

The force required to just start to slide the book is due to the static friction force.

Let N be the normal reaction

N = mg = 1.7 x 9.8 = 16.66 N

According to the laws of friction

F = μs x N

2.6 = μs x 16.66

μs = 0.156

According to the laws of friction

F' = μk x N

1.5 = μk x 16.66

μk = 0.09