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Lv 7
? asked in Science & MathematicsPhysics · 9 years ago

Urgent! Help on this AP Physics problem?

A 140 kg hoop rolls along a horizontal floor so that the hoop's center of mass has a speed of 0.150 m/s. How much work must be done on the hoop to stop it?

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  • JullyWum
    Lv 7
    9 years ago
    Favorite Answer

    Work required to stop the hoop = net KE (work-energy theorem)

    ♦ Net KE = linear KE(½mv²) + rotational KE (½Iω²)

    I = moment of inertia of loop (mr²), ω = ang vel (ω = v/r)

    KEn = ½mv² + ½(mr²)(v/r)² = 2(½mv²) = mv²

    = 140kg x (0.15m/s)² .. .. .. ►W = 3.15 J

  • zee_prime
    Lv 6
    9 years ago

    Total energy = rotational + linear kinetic energy. This is the amount of work needed to stop it, and it's equal to 1/2 I w^2 + 1/2 M v^2 where I= moment of inertia= whoops; insufficient information. You need to know the radius of the hoop.

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