Answer:A
Explanation:
From the question, the given parameters are given.
Mass M = 30kg
Radius r = 2 m
Coefficient of static friction μ = 0.8
Coefficient of kinetic friction μ = 0.6
Kinetic friction Fk = μ × mg
Fk = 0.6 × 30 × 9.8
Fk = 176.4 N
The force acting on the merry go round is a centripetal force F.
F = MV^2/r
This force must be greater than or equal to the kinetic friction Fk. That is,
F = Fk
F = 176.4
Substitute F , M and r into the centripetal force formula above
176.4 = (30×V^2)/2
Cross multiply
352.8 = 30V^2
V^2 = 352.8/30
V = sqrt (11.76) m/s
V = 5.24 m/s
Therefore, the maximum speed of the merry go round before the child begins to slip is sqrt (12) m/s approximately
We have that the the minimum coefficient of static friction is mathematically given as
[tex]\mu= 0.235[/tex]
Generally the equation for the Force is mathematically given as
[tex]F=n\mu\\\\Where\\\\n = mg\\\\Given\\\\Fc = m(vt2/R)\\\\Therefore\\\\\mu = vt2/gR\\\\\mu = 2.22/(9.8(2.1)) \\\\[/tex]
[tex]\mu= 0.235[/tex]
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