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Answer :
- The total kinetic energy of the system, consisting of a 50.0 kg merry-go-round with a radius of 1.50 m rotating at 20.0 rpm and an 18.0 kg child located 1.25 m from the center, is approximately 296 J.
Option C is answer.
1. Given:
- Mass of the merry-go-round (M) = 50.0 kg
- Radius of the merry-go-round (R) = 1.50 m
- Rotational speed (ω) = 20.0 rpm = 2.09 rad/s
- Mass of the child (m) = 18.0 kg
- Distance of the child from the center (r) = 1.25 m
2. Calculate the kinetic energy of the merry-go-round:
- Kinetic energy of the merry-go-round = 1/2 * M * (ω * R)^2
- Kinetic energy of the merry-go-round = 1/2 * 50.0 kg * (2.09 rad/s * 1.50 m)^2
- Kinetic energy of the merry-go-round ≈ 274 J
3. Calculate the kinetic energy of the child:
- Kinetic energy of the child = 1/2 * m * (ω * r)^2
- Kinetic energy of the child = 1/2 * 18.0 kg * (2.09 rad/s * 1.25 m)^2
- Kinetic energy of the child ≈ 22 J
4. Calculate the total kinetic energy of the system:
- Total kinetic energy = Kinetic energy of the merry-go-round + Kinetic energy of the child
- Total kinetic energy ≈ 274 J + 22 J
- Total kinetic energy ≈ 296 J
The total kinetic energy of the system is approximately 296 J. Option C is answer.
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