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

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 $v^2-u^2=2 as$

$1.72^2-0=2\times a\times 0.98$

$a=\frac{1.72^2}{2\times 0.98}$

$a=1.509 m/s^2$

and acceleration is also given by

$a=\frac{F-f_r}{m}$

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

F=applied Force

$a=\frac{24-\mu mg}{m}$

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

equating a

$1.509=6.85-\mu g$

$\mu g=6.85-1.509$

$\mu =\frac{5.348}{9.8}$

$\mu =0.54$