Respuesta :
Answer:
μ=0.16
Explanation:
Given that
m= 3.5 Kg
d= 0.78 m
F= 26 N
v= 1.68 m/s
Lets take coefficient of kinetic friction = μ
Friction force Fr=μ m g
Lets take acceleration of block is a m/s²
F- Fr = m a
26 - μ x 3.5 x 10 = 3.5 a ( take g =10 m/s²)
a= 7.42 - 35μ m/s²
The final speed of the block is v
v= 1.68 m/s
We know that
v²= u²+ 2 a d
u= 0 m/s given that
1.68² = 2 x a x 0.78
a= 1.80 m/s²
a= 7.42 - 35μ m/s²
7.42 - 35μ = 1.80
μ=0.16
Answer:[tex]\mu =0.573[/tex]
Explanation:
Given
mass of book [tex]m =3.5 kg[/tex]
distance moved by book [tex]s=0.78 m[/tex]
Force applied [tex]F=26 N[/tex]
initial speed [tex]u=1.68 m/s[/tex]
coefficient of kinetic friction is [tex]\mu [/tex]
using [tex]v^2-u^2=2 as[/tex]
[tex]a=\frac{0-1.68^2}{2\times 0.78}[/tex]
[tex]a=1.809 m/s^2[/tex] deceleration
Net Force
[tex]F-f_r=ma[/tex]
[tex]F-ma=f_r[/tex] , where [tex]f_r=friction\ force[/tex]
[tex]f_r=\mu N=\mu mg[/tex]
[tex]26-3.5\times 1.809=\mu \cdot 3.5\times 9.8[/tex]
[tex]\mu =\frac{26-6.33}{34.3}[/tex]
[tex]\mu =\frac{19.67}{34.3}=0.573[/tex]
