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Dahasolnce [82]
3 years ago
15

A merry-go-round makes one complete revolution in 5.3 s. A 48.5 kg child sits on the horizontal floor of the merry-go-round 2.7

m from the center. Find the horizontal force of friction that acts on the child. Answer in units of N.
Physics
1 answer:
Marianna [84]3 years ago
4 0

Answer:183.94 N

Explanation:

Given

Time Period T=5.3 s

mass of child m=48.5 kg

Radius r=2.7 m

Velocity v=\frac{2\pi \cdot r}{T}

v=\frac{2\pi \cdot 2.7}{5.3}

v=3.20 m/s

Now Centripetal Force will be Balanced by Frictional Force

Centripetal Force F_c=m\cdot \frac{v^2}{r}

F_c=48.5\cdot \frac{3.2^2}{2.7}

F_c=183.94 N

therefore Friction Force is 183.94 N

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