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 0.98 m by the horizontal 24 N force from the broom and then has a speed of 1.72 m/s, what is the coefficient of kinetic friction between the book and floor?

Answer 1

Answer:0.54

Step-by-step explanation:

Given

mass of book
`m=3.5 kg`

distance moved
`s=0.98 m`

Force applied
`F=24 N`

Speed after traveling 0.98 m is
`v=1.72 m/s`

let
`\mu`be the coefficient of kinetic Friction

using
`v^2-u^2=2 as`

`1.72^2-0=2* a* 0.98`

`a=(1.72^2)/(2* 0.98)`

`a=1.509 m/s^2`

and acceleration is also given by

`a=(F-f_r)/(m)`

where
`f_r=Frictional\ Force=\mu mg`

F=applied Force

`a=(24-\mu mg)/(m)`

`a=frac{24}{3.5}-\mu g`

equating a

`1.509=6.85-\mu g`

`\mu g=6.85-1.509`

`\mu =(5.348)/(9.8)`

`\mu =0.54`