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17 August, 20:17

The 0.5-kg collar C starts from rest at A and slides with negligible friction on the fixed rod in the vertical plane. Determine the velocity v with which the collar strikes end B when acted upon by the 5-N force, which is constant in direction. Neglect the small dimensions of the collar. [Use Principle of energy conservation]

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  1. 17 August, 20:35
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    The answer is 2.32 m/s

    Explanation:

    Solution

    Given that:

    U a-b = ΔT

    Thus

    F cos 30° * hₐ - F sin 30° * hₐ + Whₐ = 1/2 m (v²b - v²ₐ)

    Now,

    vb = √2Fhₐ (cos 30° - sin 30°) + mghₐ / m + v²ₐ

    Where

    F = This is the force acting on the collar

    m = this is the mass on the collar C

    g = The acceleration due to gravity

    hₐ = The height of the collar at the position A

    vb = The velocity of the collar at position B

    vₐ = The velocity of the collar at A

    So,

    We replace 5N for F, 0.2 m for hₐ, 0.5 kg for m, 9.81 m/s for g, 0 for vₐ

    Now,

    vb = √ 2 * (5 * 0.2 * (cos 30° - sin 30°) + 0.5 * 9.81 * 0.2 / 0.5 + 0

    =√2.694/0.5

    =2.32 m/s

    Hence, the he velocity v with which the collar strikes the end B is 2.32 m/s
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