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 0.91 m by the horizontal 25 N force from the broom and then has a speed of 1.53 m/s, what is the coefficient of kinetic friction between the book and floor?

Answer 1

The coefficient of kinetic is

`u_(k)=0.59`

Step-by-step explanation:

The forces in the axis 'x' and 'y' using law of Newton to find coefficient of kinetic friction

ΣF=m*a

ΣFy=W-N=0

ΣFy=Fn-Fu=m*a

`F_(u) =u_(k) *N\\F_(N)=25N\\N=W\\N=3.5kg*9.8(m)/(s^(2) )=34.3N`

`F_(N)-F_(u)=m*a\\F_(N)-u_(k)*N=m*a\\u_(k)*N=F_(N)-m*a\\u_(k)=(F_(N)-m*a)/(N)`

Now to find the coefficient can find the acceleration using equation of uniform motion accelerated

`v_(f) ^(2)=v_(o)^(2)+2*a(x_(f)-x_(o))\\x_(o)=0\\v_(o)=0\\v_(f) ^(2)=2*a*x_(f)\\a=(v_(f) ^(2))/(2*a*x_(f))\\ a=((1.53(m)/(s) )^(2))/(2*0.91m)\\a= 1.28 (m)/(s^(2) )`

So replacing the acceleration can fin the coefficient:

`u_(k)=(F_(N)-m*a )/(N)\\u_(k)=(25N-(3.5kg*1.28(m)/(s^(2)) )/(34.3N) \\u_(k)=0.59`