A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a distance d. W

Question

A block of mass m=1.2 kg is held at rest against a spring with a force constant k=730N/m. Initially the spring is compressed a distance d. When the block is released, it slides across a surface that is frictionless except for a rough patch of width ????x= 5.0 cm that has coefficient of kinetic friction µk =0.44 Find a general expression of d such that the block’s speed after crossing the rough patch is Vf = 2.3 m/s. Also calculate the value of d. How does the friction effect the d to reach the same Vf?

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Ngọc Diệp 5 years 2021-07-28T18:38:51+00:00 1 Answers 60 views 0

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  1. linhdan
    0
    2021-07-28T18:40:19+00:00
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    Answer:

    Explanation:

    potential energy of compressed spring

    = 1/2 k d²

    = 1/2 x 730 d²

    = 365 d²

    This energy will be given to block of mass of 1.2 kg in the form of kinetic energy .

    Kinetic energy after crossing the rough patch

    = 1/2 x 1.2 x 2.3²

    = 3.174 J

    Loss of energy

    = 365 d² – 3.174  

    This loss is due to negative work done by frictional force

    work done by friction = friction force x width of patch

    = μmg d ,   μ = coefficient of friction , m is mass of block , d is width of patch

    = .44 x 1.2 x 9.8 x .05

    = .2587 J

    365 d² – 3.174   = .2587

    365 d² = 3.4327

    d² = 3.4327 / 365

    = .0094

    d = .097 m

    = 9.7 cm

    If friction increases , loss of energy increases . so to achieve same kinetic energy , d will have to be increased so that initial energy increases so compensate increased loss .

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