a 30 kg child is sitting 2 meters from the center of a merry go round. The coefficients of static and kinetic friction between the child and the surface of the merry go round are 0.8 and 0.6 respectively. Determine the maximum speed of the merry go round before the child begins to slip. A) m/s B) 4 m/s c) 8 m/s D) 12 m/s E)16 m/s

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

an energy transformation flow diagram is shown

where is it ?

radiation

explanation:

heat is transmitted by radiation