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V125BC [204]
3 years ago
12

Two hoops, starting from rest, roll down identical inclined planes. The work done by nonconservative forces, such as air resiste

nce, is zero (Wnc = 0J). Both have the same mass M, but, one hoop has the twice the radius of the other. The moment of inertia for each hoop is I = Mr sequared, where r is the radius. Which hoop, if either has the greater total kinetic energy (transational and rotational) at the bottom of the incline.
Hoop 1 Radius = R and Mass = M
Hoop 2 Radius = 1/2 R and Mass = M
Physics
1 answer:
Maurinko [17]3 years ago
6 0

Answer: both hoops have the same kinetic energy at the bottom of the incline.

Explanation:

If we assume no work done by non conservative forces (like friction) , the total mechanical energy must be conserved.

K1 + U1 = K2 + U2

If both hoops start from rest, and we choose the bottom of the incline to be the the zero reference level for gravitational potential energy, then

K1 = 0 and U2 = 0

⇒ ΔK = ΔU = m g. h

If both inclines have the same height, and both hoops have the same mass m, the  change in kinetic energy, must be the same for both hoops.

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Option C is the correct answer.

Explanation:

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Slav-nsk [51]

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The correct option that gives the relationships between the forces is the the option;

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<em />

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Force exerted on the ball by the bat = F_{bat}

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\underline{F_{ball}}<u> on bat</u> = \underline{F_{bat}}<u> on ball</u>

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Learn more Newton's Laws of motion here:

brainly.com/question/24522313

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