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 0.75 m by the horizontal 25 n force from the broom and then has a speed of 1.83 m/s, what is the coefficient of kinetic friction between the book and floor?

0.6

Explanation:

from the question we are given the following:

mass of the book (m) = 3.5 kg

distance (s) = 0.75 m

horizontal force (Fh) = 25 N

final velocity (V) = 1.83 m/s

acceleration due to gravity (g) = 9.8 m/s^{2}

initial velocity (u) = 0 ( since the book was initially at rest)

coefficient of friction ( Uk) = ?

We can get the coefficient of friction from the equation

net force = horizontal force - frictional force

where net force = m x a  

frictional force = Uk x m x g

the equation now becomes

m x a = Fh - (Uk x m x g)

but we need to find the acceleration to apply the equation, and we can get the acceleration from V^{2} = U^{2} + 2as

therefore

1.83 = 0 + (2 x a x 0.75)

a = 1.22 m/s^{2}

substituting the value of "a" into our equation we have

3.5 x 1.22 = 25 - (Uk x 3.5 x 9.8)

Uk = 0.6

answ675857-233333

explanation:

at positions a, g kinetic energy is maximum.

at position d, potential energy is maximum.