Posted by Help! on .
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 0.88 m by the horizontal 25 N force from the broom and then has a speed of 1.63 m/s, what is the coefficient of kinetic friction between the book and floor?

college physics 
drwls,
Work done by broom = Kinetic energy increase + work done against fraction
Compute the friction work using that equation. Divide it by the distance moved to get the friction force. Use that to get the friction coefficient 
college physics 
lilly,
m = 3.5kg
x = 0.88m
f = 25N
speed = 1.63m/s
coefficient of kinetic friction = ?
we know, fk=fnormal*coefficient of kinetic friction
where fk = kinetic frictional force
so, fk= mg*coefficient
fk=3.5*9.8*cefficient
a=?
so applying the equation
v^2=u^2+2ax
a= 1.1m/s^2
fnet=ma
so, ma=3.5*9.8*coefficient
cefficient= 0.11