3/122 The collar has a mass of 2 kg and is attached to the light spring, which has a stiffness of 30 N/m and an unstretched length of 1.5 m.

Question

3/122 The collar has a mass of 2 kg and is attached to the light spring, which has a stiffness of 30 N/m and an unstretched length of 1.5 m. The collar is released from rest at A and slides up the smooth rod under the action of the constant 50‐N force. Calculate the velocity v of the collar as it passes position B.

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Thông Đạt 5 years 2021-09-04T04:47:04+00:00 1 Answers 365 views 1

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  1. kimcuc
    1
    2021-09-04T04:48:25+00:00
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    Explanation:

    According to the law of conservation of energy, work done by the force is as follows.

            W_{F} = F Cos (30^{o}) \times 1.5

                       = 64.95 J

    Now, gain in potential energy is as follows.

                    P.E = mgh

                          = 2 \times 9.8 \times 1.5

                          = 29.4 J

    Gain in potential energy will be as follows.

               = \frac{1}{2}kx^{2}_{2} - \frac{1}{2}kx^{2}_{1}

               = \frac{1}{2} \times 30 N/m \times [(2.5 - 1.5)^{2} - (2 - 1.5)^{2}]

               = 11.25

    As,

              W_{f} = u_{1} + u_{2} + \frac{1}{2}mv^{2}

              \frac{1}{2}mv^{2} = W_{f} - u_{1} - u_{2}  

                       = 64.95 J – 29.4 – 11.25

                       = 24.3

                  v^{2} = \frac{24.3 \times 2}{2}

                      v = 4.92 m/s

    Therefore, we can conclude that relative velocity at point B is 4.92 m/s.  

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