Question

When you push a 1.90-kg book resting on a tabletop, you have to exert a force of 2.10 N to start the book sliding. Once it is sliding, however, you can use a force of only 1.06 N to keep the book moving with constant speed. Part A What is the coefficient of static friction between the book and the tabletop?

Answer 1

static frictional force = ƒs = (Cs) • (Fɴ)
              2.26 = (Cs) • m • g
              2.26 = (Cs) • (1.85) • (9.8)
           Cs = 0.125 
kinetic frictional force = ƒκ = (Cκ) • (Fɴ)
              1.49 = (Cκ) • m • g
              1.49 = (Cκ) • (1.85) • (9.8)
           Cκ = 0.0822 

Answer 2

When you push a 1.90-kg book resting on a tabletop with a force of 2.10 N, the coefficient of static friction between the book and the tabletop is 0.11.  

A) When the book is resting on a tabletop and then is pushed with a force of 2.10 N, we have:

$F - F_{\mu_{s}} = ma$   (1)

Where:

F: is the applied force = 2.10 N

$F_{\mu_{s}}$: is the force due to the coefficient of static friction = $\mu_{s}N$  

m: is the mass of the book = 1.90 kg

a: is the acceleration

N: is the normal force = mg

g: is the acceleration due to gravity = 9.81 m/s²

$\mu_{s}$: is the coefficient of static friction =?

Since the book is at rest, the acceleration is zero, so equation (1) is:

$F - F_{\mu_{s}} = 0$    

$F - \mu_{s}N = 0$    

$F - \mu_{s}mg = 0$    

Solving the above equation for $\mu_{s}$ we have:

$2.10 N - \mu_{s}*1.90 kg*9.81 m/s^{2} = 0$    

$\mu_{s} = \frac{2.10 N}{1.90 kg*9.81 m/s^{2}} = 0.11$

Therefore, the coefficient of static friction is 0.11.

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I hope it helps you!