Respuesta :
Answer:
uk = 0.25
Explanation:
Given:-
- An object comes to stop with acceleration, a = -2.45 m/s^2
Find:-
What is the coefficient of kinetic friction between the object and the floor?
Solution:-
- Assuming the object has mass (m) that slides over a rough surface with coefficient of kinetic friction (uk). There is only Frictional force (Ff) acting in the horizontal axis on the object opposing the motion (-x direction).
- We will apply equilibrium equation on the object in vertical direction.
N - m*g = 0
N = m*g
Where, N : Contact force exerted by the surface on the floor
g : Gravitational acceleration constant = 9.81 m/s^2
- Now apply Newton's second law of motion in the horizontal ( x-direction ):
- Ff = m*a
- The frictional force is related to contact force (N) by the following expression:
Ff = uk*N
- Substitute the 1st and 3rd expressions in the 2nd equation:
uk*m*g = -m*a
uk = a / g
- Plug in the values and solve for uk:
uk = - (-2.45) / 9.81
uk = 0.25
The coefficient of kinetic friction is 0.25
Given to us,
constant deacceleration = -2.45 m/sec²
A normal force is a contact force that is been applied between two objects.
In this case, the surface is applying the normal force on the object, which is opposite to the gravitational force.
[tex]\bold{Normal force = Mass\times acceleration\ due\ to\ gravity}[/tex]
[tex]\bold{N=m\times g}[/tex]
The friction force is the force that is been exerted by the surface on the object, as an object moves across it or makes an effort to move across it.
[tex]\bold{ Frictional\ force= coefficient\ of\ kinetic\ friction \times Normal\ force}[/tex]
[tex]\bold{ Frictional\ force= \mu \times Normal\ force}[/tex]
Also, we know that according to Newton's law, the force is the dot product of mass and acceleration. And, in this case, a force is also been applied that caused the object to move in the initial case.
[tex]\bold{ Force= Mass \times acceleration}[/tex]
[tex]\bold{ Force= m \times a}[/tex]
Calculating the net force on the horizontal plane,
[tex]\bold{Frictional\ Force= Force}[/tex]
[tex]\bold{\mu_k \times N= m \times a}[/tex]
[tex]\mu_k \times m \times g = m \times a[/tex]
canceling the mass(m) from both sides,
[tex]\mu_k \times m \times g = m \times a\\\mu_k \times g = a\\\mu_k = \dfrac{a}{g}\\\\\mu_k = \dfrac{-2.45}{9.81}\\\\ \mu_k=0.2497 = 0.25[/tex]
Note: the coefficient of friction can not be negative.
Hence, the coefficient of kinetic friction is 0.25
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