Question

In a pickup game of dorm shuffleboard,students crazed by final exams use a broom to propel a calculus book along the dorm hallway. If the 3.5 kg book is pushed from rest through a distance of 0.90 m by the horizontal 25 N force from the broom and then has a speed of 1.60 m/s,what is the coefficient of kinetic friction between the book and floor?

Answer 1

Answer:

Coefficient of friction between the book and floor is 0.582.

Explanation:

Using the velocity formula;

v^2 = 2as

a = v^2/(2s)

a = 1.6^2/(2*0.9)

a = 2.56/1.8

a = 1.42 m/s^2

the force necessary to give the book the acceleration is  

F = ma = 3.5*1.42 (m is mass of the book i.e. 3.5 kg)

F = 4.98 N

The difference in the force is the friction force, which is

Ff = 25 - 4.98 = 20 N

Ff = mgμ

where μ is coefficient of friction and g is acceleration due to gravity that is 9.8 m/s^2

μ = Ff/mg

μ = 20/(3.5*9.81)

μ = 0.582

Coefficient of friction between the book and floor is 0.582.

Answer 2

This question involves the concepts of the equation of motion and Newton's second law of motion.

The coefficient of kinetic friction between the book and the floor is "".

First, we will find the acceleration of the block by using the third equation of motion:

$2as = v_f^2-v_i^2$

where,

a = acceleration = ?

s = distance covered = 0.9 m

vf = final speed = 1.6 m/s

vi = initial speed = 0 m/s

Therefore

$a=\frac{(1.6\ m/s)^2-(0\ m/s)^2}{2(0.9\ m)}\\\\a=1.42\ m/s^2$

Hence, from Newton's second law of motion:

$Net\ Force = Frictional\ Force + F\\Net\ Force = \mu mg+ma$

where,

Net Force = 25 N

μ = coefficient of kinetic friction = ?

m = mass of the book = 3.5 kg

g = acceleration due to gravity = 9.81 m/s²

Therefore,

$25\ N = \mu(3.5\ kg)(9.81\ m/s^2)+(3.5\ kg)(1.42\ m/s^2)\\\\\mu=\frac{25\ N - 4.98\ N}{34.33\ N}\\\\\mu = 0.58$

Learn more about Newton's Second Law of Motion here:

brainly.com/question/13447525?referrer=searchResults

The attached picture shows Newton's Second Law of Motion.