4. . When you push a 1.80-kg book resting on a tabletop, it takes 2.25 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?

The coefficient of static friction is

The coefficient of kinetic friction is

Explanation:

From the question we are told that

   The mass of the book is  m =  1.80 kg

    The force required by the book to start sliding is

    The force required to keep the booking moving at constant speed is

Generally the static  frictional force is mathematically represented as  

This static frictional  force is equivalent to the force required for the book to start sliding

So

=>      

=>      

Generally the kinetic  frictional force is mathematically represented as  

This static frictional  force is equivalent to the force required for the book to  slide at constant velocity

So

=>      

=>      

=>      

the simple machine has three main features. they are:

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever

the answer would be   dignified - apex