A 1.150 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to displace the

Question

A 1.150 kg air-track glider is attached to each end of the track by two coil springs. It takes a horizontal force of 0.900 N to displace the glider to a new equilibrium position, x= 0.170 m. Find the effective spring constant of the system. Submit Answer Tries 0/12 The glider is now released from rest at x= 0.170 m. Find the maximum x-acceleration of the glider. Submit Answer Tries 0/12 Find the x-coordinate of the glider at time t= 0.470T, where T is the period of the oscillation. Submit Answer Tries 0/12 Find the kinetic energy of the glider at x=0.00 m.

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Edana Edana 4 years 2021-09-05T11:43:13+00:00 1 Answers 1 views 0

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  1. Delwyn
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    2021-09-05T11:44:33+00:00
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    k = 5.29

    a = 0.78m/s²

    KE = 0.0765J

    Explanation:

    Given-

    Mass of air tracker, m = 1.15kg

    Force, F = 0.9N

    distance, x = 0.17m

    (a) Effective spring constant, k = ?

    Force = kx

    0.9 = k X0.17

    k = 5.29

    (b) Maximum acceleration, m = ?

    We know,

    Force = ma

    0.9N = 1.15 X a

    a = 0.78 m/s²

    c) kinetic energy, KE of the glider at x = 0.00 m.

    The work done as the glider was moved = Average force * distance

    This work is converted into kinetic energy when the block is released. The maximum kinetic energy occurs when the glider has moved 0.17m back to position x = 0  

    As the glider is moved 0.17m, the average force = ½ * (0 + 0.9)

    Work = Kinetic energy

    KE = 0.450 * 0.17

    KE = 0.0765J

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