A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top of a relaxe

Question

A metal block of mass 3 kg is falling downward and has velocity of 0.44 m/s when it is 0.8 m above the floor. It strikes the top of a relaxed vertical spring of length 0.4 m. The spring constant is 2000 N/m. After striking the spring, the block rebounds. What is the maximum height above the floor that the block reaches after the impact

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Ngọc Khuê 6 months 2021-07-20T21:42:46+00:00 2 Answers 5 views 0

Answers ( )

    0
    2021-07-20T21:43:59+00:00

    Answer:

    y_{max} = 0.829\,m

    Explanation:

    Let assume that one end of the spring is attached to the ground. The speed of the metal block when hits the relaxed vertical spring is:

    v = \sqrt{(0.8\,\frac{m}{s})^{2} + 2\cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m)}

    v = 2.913\,\frac{m}{s}

    The maximum compression of the spring is calculated by using the Principle of Energy Conservation:

    (3\,kg)\cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m) + \frac{1}{2}\cdot (3\,kg)\cdot (2.913\,\frac{m}{s} )^{2} = (3\,kg) \cdot (9.807\,\frac{m}{s^{2}})\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s) ^{2}

    After some algebraic handling, a second-order polynomial is formed:

    12.728\,J = \frac{1}{2}\cdot (2000\,\frac{N}{m} )\cdot (\Delta s)^{2} - (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \Delta s

    1000\cdot (\Delta s)^{2}-29.421\cdot \Delta s - 12.728 = 0

    The roots of the polynomial are, respectively:

    \Delta s_{1} \approx 0.128\,m

    \Delta s_{2} \approx -0.099\,m

    The first root is the only solution that is physically reasonable. Then, the elongation of the spring is:

    \Delta s \approx 0.128\,m

    The maximum height that the block reaches after rebound is:

    (3\,kg) \cdot (9.807\,\frac{m}{s^{2}} )\cdot (0.4\,m-\Delta s) + \frac{1}{2}\cdot (2000\,\frac{N}{m})\cdot (\Delta s)^{2} = (3\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot y_{max}

    y_{max} = 0.829\,m

    0
    2021-07-20T21:44:36+00:00

    Answer:

    0.81 m

    Explanation:

    In all moment, the total energy is constant:

    Energy of sistem = kinetics energy + potencial energy = CONSTANT

    So, it doesn’t matter what happens when the block hit the spring, what matters are the (1) and (2) states:

    (1): metal block to 0.8 m above the floor

    (2): metal block above the floor, with zero velocity ( how high, is the X)

    Then:

    E_{kb1} + E_{gb1}  = E_{kb2} + E_{gs2}

    E_{kb1} + E_{gb1}  = 0 + E_{gs2}

    \frac{1}{2}*m*V_{b1} ^{2}   + m*g*H_{b1}  = m*g*H_{b2}

    H_{b2}  =  \frac{V_{b1} ^{2} }{2g}  + H_{b1}

    Replacing data:

    H_{b2}  =  \frac{0.44^{2} }{2*9.81}  + 0.8

    H_{b2}  =  \frac{0.44^{2} }{2*9.81}  + 0.4

    HB2 ≈ 0.81 m

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