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 0.82 m by the horizontal 24 N force from the broom and then has a speed of 1.40 m/s, what is the coefficient of kinetic friction between the book and floor

Answer 1

`\mu_(k) = 0.578 N`

Step-by-step explanation:

given

mass = 3.5 kg

distance d = 0.82 m

horizontal force = 24 N

coefficient of kinetic friction between the book and floor
`\mu _(k)` = ?

now by using Newton ' s second law

`\Sigma f = ma`

for vertical motion

W- N = 0 ( where w is weight and N is normal reaction)

W = N = mg .....(i) ( m is mass of book and g is acceleration due to gravity )

for horizontal motion

`f_(h)- \mu_(k)N = ma .............(ii)`

so
`\mu_(k) =( f_(h)- ma)/(N)` ..........(iii)

from (i)

`N = mg = 3.5 * 9.8 = 34.3 N`

for acceleration a using kinematics equations

`v^2 - u^2 = 2ad`

`a =( v^2-u^2)/(2d)`

`a = ((1.40)^2 - 0^2)/( 2 * 0.82)\\a= 1.19 m/s^2`

now substitute the values in equation (iii) we get

`\mu_(k) =(24- 3.5 * 1.19)/(34.3)\\\\\mu_(k) = 0.578`