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Height of the rise of the object attached to the spring ?

  1. Oct 31, 2016 #1
    1. The problem statement, all variables and given/known data
    Question :- A block of mass ##5 kg## is attached to a spring. The spring is stretched by ##10 cm## under the load of the block. A impulse is provided to the block such that it moves up with a velocity of ##2 m/s##. Find the height it will rise.

    sdasd.png
    2. Relevant equations
    ##F_s = kx##
    ##x = 0.1m##
    ##v = 2m/s##
    ##m = 5kg##
    ##x_2## - height raised from the mean position of the spring.

    3. The attempt at a solution
    We can get spring constant by equating spring force and weight of the object.
    $$kx = mg $$
    $$k = {5 \times 10 \over 1/10} = 500 \hspace{20 mm} (1)$$

    Now initial total energy of the system :-

    $$TE_i = {m v^2 + kx^2\over 2}$$
    Now using given data and from ##(1)##.
    $$TE_i = {5 \times 4 + 500\times 1/100\over 2} = {25 \over 2}$$

    final total energy :-

    Velocity at maximum will be zero.

    $$TE_f = {mg(x + x_2) + kx_2^2\over 2} \hspace{20 mm} (2)$$

    Using conservation of energy ,##(1)## and ##(2)## we get,

    $${25 \over 2} = {mg(x + x_2) + kx_2^2\over 2}$$
    $$20x_2^2 + 2x_2 -0.8 = 0$$

    ##x_2 = 0.15##
    thus total height is ##0.25 m##.
    The answer is ##0.20 m##. I think the answer is incorrect, since this is pretty straight forward question.
     

    Attached Files:

  2. jcsd
  3. Oct 31, 2016 #2

    Doc Al

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    Staff: Mentor

    In Equation (2) you divided the gravitational PE by 2.
     
  4. Oct 31, 2016 #3
    Such a silly mistake and i am trying this question for more than an hour.:frown::frown::frown::mad::mad::mad:
     
    Last edited: Oct 31, 2016
  5. Oct 31, 2016 #4

    Doc Al

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    Staff: Mentor

    That's the story of my life! :smile:
     
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