A block of ice of mass 4.10 kg is placed against a horizontal spring that has force constant k = 210 N/m and is compressed a distance 2.60×1

Question

A block of ice of mass 4.10 kg is placed against a horizontal spring that has force constant k = 210 N/m and is compressed a distance 2.60×10−2 m. The spring is released and accelerates the block along a horizontal surface. You can ignore friction and the mass of the spring.

Calculate the work done on the block by the spring during the motion of the block from its initial position to where the spring has returned to its uncompressed length.

What is the speed of the block after it leaves the spring?

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Thái Dương 3 years 2021-08-15T19:47:40+00:00 1 Answers 29 views 0

Answers ( )

  1. Doris
    0
    2021-08-15T19:48:42+00:00
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    Answer:

    (i) Work done on the block will be 0.071 J.

    (ii) Speed of the block will be  0.19 m/s.

    Explanation:

    Since We know that Potential energy stored in the spring when it is compressed to some distance is given by:-

    P. E = \frac{1}{2} k.x²

    where k = Spring constant

               x = compression of spring

    Thus, P.E = \frac{1}{2} × 210 × (2.60 × 10⁻²)²

                = 0.071 J.

    (i) Mow the work done on the block will bw equal to the potential energy stored in the spring.

    Thus, W = P.E

     or, W = 0.071 J.

    (ii) The potential energy of the spring will get converted into the kinrtic energy of the block.

    Let K.E = \frac{1}{2} mv²    where v = speed of the block

                                             m = mass of the block.

    Now K.E = P.E

    or, \frac{1}{2} mv²  = 0.071

     or, v² =   \dfrac{0.071 \times 2}{m}

      or, v^{2} =    \dfrac{0.071 \times 2}{4.10}

       or, v^{2}  = 0.0346

     or, v = \sqrt{0.0346}

      or, v = 0.19 m/s

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