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6 August, 09:49

A uniform solid disk is released from rest and rolls without slipping down an inclined plane that makes an angle of 25° with the horizontal. What is the forward speed of the disk after it has rolled 3.0 m, measured along the plane?

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  1. 6 August, 09:56
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    The forward speed of the disk is 4.07 m/s

    Explanation:

    Step 1: Data given

    Angle = 25 °C

    Distance the disk has rolled = 3.0 m

    Step 2 = Calculate the initial energy

    Since the disk is, at moment t=0, at rest and at a height we only have potential energy

    Ei = m*g*h

    ⇒with E = the potential energy

    ⇒with m = the mass of the disk

    ⇒with g = 9.81 m/s²

    ⇒with h = the height = x*sin 25° = 3*sin25°

    Ei = m*g*X*sin25

    Step 3: Calculate the final energy

    Ef = 1/2 (mv²) + 1/2 Iw²

    Ef = 1/2 (mv²) + 1/2 (mR²) (v/R) ²

    Ef = 1/2mv² + 1/4 mv² = 3/4 mv²

    Step 4: The law of conservation of energy

    Ei = Ef

    m*g*X*sin25 = 3/4 mv²

    g*X*sin25 = 3/4 v²

    Step 5: Calculate v

    v = √ (4/3 * g*X sin25°)

    v = √ (4/3 * 9.81*3 sin25°)

    v = 4.07 m/s

    The forward speed of the disk is 4.07 m/s
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