Respuesta :
F = ma
where
F = frictional force
m = mass of the block = 1.4 kg (given)
a = acceleration of the block = 1.25 m/sec^2 (given)
Substituting values,
F = (1.4)(1.25)
F = 1.75 N
By definition,
F = (mu)(Normal force)
where
mu = coefficient of friction
Normal force = mg = 1.4*9.8 = 13.72
Again, substituting appropriate values,
1.75 = mu(13.72)
mu = 0.128
where
F = frictional force
m = mass of the block = 1.4 kg (given)
a = acceleration of the block = 1.25 m/sec^2 (given)
Substituting values,
F = (1.4)(1.25)
F = 1.75 N
By definition,
F = (mu)(Normal force)
where
mu = coefficient of friction
Normal force = mg = 1.4*9.8 = 13.72
Again, substituting appropriate values,
1.75 = mu(13.72)
mu = 0.128
The coefficient of kinetic friction for the block k slides across a rough surface is 0.128.
What is Newtons second law of motion?
Newtons second law of motion shows the relation between the force mass and acceleration of a body.
It says, that the force applied on the body is equal to the product of mass of the body and the acceleration of it.
It can be given as,
[tex]F=ma[/tex]
Here, (m) is the mass of the body and (a) is the acceleration.
- Step 1) Find the friction force-
As the mass of the block is 1.4 kg and the acceleration of it is 1.25 m/s (squared). Therefore the friction force can be given as using the above formula as,
[tex]F_f=1.4\times1.25\\F_f=1.75\rm N[/tex]
- Step 2) Find the normal force-
The normal force acting on it due to gravity force is,
[tex]F_n=1.4\times9.8\\F_n=13.72\rm N[/tex].
- Step 3) Find the coefficient of kinetic friction[tex](\mu)[/tex]-
Now the friction force acting on a moving body is equal to the product of normal force acting on it and the coefficient of kinetic friction.
Thus, the coefficient of kinetic friction[tex](\mu)[/tex] can be given as,
[tex]F_f=F_N\times\mu\\1.75=13.72\times \mu\\\mu=0.128[/tex]
Hence, the coefficient of kinetic friction for the block k slides across a rough surface is 0.128.
Learn more about the Newtons second law of motion here;
https://brainly.com/question/25545050