Answer:
The coefficient of kinetic friction between the cart and the track is 0.114
Explanation:
Given;
Angle of inclination, Īø = 10Ā°
Acceleration of the cart, a = 0.60 m/sĀ²
Apply Newton's law of motion
mgsinĪø - Ī¼kmgcosĪø = ma
Divide through by mass, m
gsinĪø - Ī¼kgcosĪø = a
Ī¼kgcosĪø = gsinĪø - a
[tex]\mu _k = \frac{gsin \theta -a}{gcos \theta}[/tex]
where;
Ī¼k is the coefficient of kinetic friction between the cart and the track
Substitute the given values and calculate coefficient of kinetic friction Ī¼k
[tex]\mu _k = \frac{gsin \theta -a}{gcos \theta} \\\\\mu _k = \frac{9.8sin 10 -0.6}{9.8cos 10}\\\\\mu _k = 0.114[/tex]
Therefore, the coefficient of kinetic friction between the cart and the track is 0.114
Answer:
The kinetic friction between the cart and the track is 0.11
Explanation:
The force at equilibrium is equal to:
F = FĀ“ - Fk
F = m*a
FĀ“ = m*g*sinĪøā
Fk = uk*m*g*cosĪøā
Replacing:
m*a = m*g*sinĪøā - uk*m*g*cosĪøā
a = g(sinĪøā - uk*cosĪøā)
Where a = 0.6 m/sĀ²
g = 9.8 m/sĀ²
Īøā = 10Ā°
Replacing values:
0.6 = 9.8(sin(10) - uk*cos(10))
Clearing uk:
uk = 0.11
