A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The spring has a forc

Question

A block of mass 5.6 kg is attached to a horizontal spring on a rough floor. Initially the spring is compressed 3.5 cm. The spring has a force constant of 1040 N/m. The coefficient of kinetic friction between the block and the floor is 0.26. The block is released from rest. The speed of the block after it has traveled 0.020 m is

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Jezebel 6 months 2021-07-29T10:07:30+00:00 1 Answers 11 views 0

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    2021-07-29T10:09:26+00:00

    Answer:

    The velocity of block = 0.188 \frac{m}{s}

    Explanation:

    Mass m = 5.6 kg

    k = 1040 \frac{N}{m}

    \mu = 0.26

    x_{1} = 0.035 m  , v_{1} = 0

    x_{2} = 0.02 m

    From work energy theorem

    K_{1} + U_{1} + W_{other} = K_{2} + U_{2}  ——— (1)

    Kinetic energy

    K = \frac{1}{2} k x^{2}  ——- (1)

    Potential energy

    U = \frac{1}{2} k x^{2} ——- (2)

    Work done

    W = F.s —— (3)

    From Newton’s second law

    R_{N} = mg

    R_{N}  = 5.6 × 9.81 = 54.9 N

    Friction  force = 0.4 × 54.9 = 21.9 N

    Now the work done by the friction

    W_{f} = – 21.9 × 0.015

    W_{f} = – 0.329 J

    Now kinetic energy

    At point 1

    K_{1} = \frac{1}{2} m v^{2} _{1}

    K_{1} = 0

    U_{1} = \frac{1}{2} k x^{2}

    U_{1} = \frac{1}{2}  (1040) 0.035^{2}

    U_{1} = 0.637 J

    At point 2

    K_{2} = \frac{1}{2} (5.6) v^{2} _{2}

    K_{2} = 2.8 v_{2} ^{2}

    Potential energy

    U_{2} = \frac{1}{2}  k x_2^{2}

    U_{2} = \frac{1}{2}  (1040) 0.02^{2}

    U_{2} = 0.208 J

    From equation (1) we get

    0 + 0.637 – 0.329 = 2.8 v_{2} ^{2} + 0.208

    2.8 v_{2} ^{2} = 0.1

    v_{2} = 0.188 \frac{m}{s}

    This is the velocity of block.

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