Respuesta :
Answer:
[tex]\mu_{k} = 0.578 N[/tex]
Explanation:
given
mass = 3.5 kg
distance d = 0.82 m
horizontal force = 24 N
coefficient of kinetic friction between the book and floor [tex]\mu _{k}[/tex] = ?
now by using Newton ' s second law
[tex]\Sigma f = ma[/tex]
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
[tex]f_{h}- \mu_{k}N = ma .............(ii)[/tex]
so [tex]\mu_{k} =\frac{ f_{h}- ma}{N}[/tex] ..........(iii)
from (i)
[tex]N = mg = 3.5 \times 9.8 = 34.3 N[/tex]
for acceleration a using kinematics equations
[tex]v^2 - u^2 = 2ad[/tex]
[tex]a =\frac{ v^2-u^2}{2d}[/tex]
[tex]a = \frac{(1.40)^2 - 0^2}{ 2 \times 0.82}\\a= 1.19 m/s^2[/tex]
now substitute the values in equation (iii) we get
[tex]\mu_{k} =\frac{24- 3.5 \times 1.19}{34.3}\\\\\mu_{k} = 0.578[/tex]
