Physics Problem: A .40 kg mass is attached to a horizontal spring with a spring constant of 50 N/m.?

A .40 kg mass is attached to a horizontal spring with a spring constant of 50 N/m. The mass is initially stretched a distance of 48 cm from its equilibrium position. The mass is then released from rest. Assume there is no friction or air resistance. Use energy of conservation to calculate the speed of the mass when the spring is compressed to a distance of 39 cm on the other side of equilibrium. Now assume there is friction and/or air resistance. You find that when the mass comes to rest for the first time after being released (the spring is now fully compressed on the other side of the equilibrium), it is compressed to a final distance of 40 cm. What types of energy change between the initial and final state and by how many joules does the internal energy increase between these times?

*****I found the speed of the mass to be 3.1 m/s but I have no idea on how to find the energy joules increase***** Thanks in advance

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  • Jim
    Lv 7
    4 years ago
    Favorite Answer

    To confirm UR answer:

    SPE at initial stretch = 1/2kx² = (0.5)(50)(0.48)² = 5.76 J {max amt of KE of mass}

    with no friction forces acting

    SPE at 39 cm compression = 1/2kx² = (0.5)(50)(0.39)² = 3.8025 J

    mass has a KE = 5.76 - 3.8025 = 1.9575 J

    speed of mass = √2KE/m = √[2(1.9575)/0.4] = √9.7875 = 3.1285 ≈ 3.13 m/s <= ans confirmed :>)

    ---

    SPE of spring at full compression on other side = 1/2kx² = 0.5(50)(0.4)² = 4.00 J

    OK... notice that initially the mas is not moving so its KE = 0 and the total (max) Mechanical Energy stored in spring = 5.76 J. So this is the MAX mechanical energy that the system is given.

    then

    after mass moves to the fullest extent (compression) on other side of spring, again the mass has stopped (temporarily) moving so again its KE = 0, here, and the spring has only got an SPE = 4.00 J. In other words there has been a LOSS of 1.76 J just by having mass move thru 1/2 a cycle (where cycle = a complete back & forth movement).

    Now the loss in total Mech Energy would occur in the spring movement, itself, as well as between the moving mass and the surface it slides on. Friction for the most part creates heat which is the form that the lost mechanical energy takes. I think it very UNspecific to ask if "THE internal energy" increases. The internal energy of WHAT??

    There are three(3) things with internal energy: 1-spring; 2-mass; 3-surface mass rides on

    I suppose one could say that the internal energy (in the form of temperature) increases {very slightly} for all three, as a general answer :>)

    • BONNIE4 years agoReport

      that helped lots, thank you

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  • 4 years ago

    For the last part:

    The initial energy (KE+PE) is 0 + ½k.xi² = ½*50*0.48²

    The final energy (KE+PE) is 0 + ½k.xf² = ½*50*0.40²

    Energy 'lost' is ½*50*(0.48² - 0.40²) = 1.76J

    1.76J has been converted to heat (this is the increase in internal energy). Ths is due to the negative work done by friction and/or air resistance.

    (This only makes sense if we include the air and any surfaces in contact with the block/spring as part of the total system. The question is not very well worded.)

    • BONNIE4 years agoReport

      that helped lots, thank you

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  • 4 years ago

    ½kx² = ½*50*0.48² = ½*50*0.39² + ½*0.4*V1²

    (50*0.48² - 50*0.39²)/0.4 = V² = 3.1²

    The energy lost = ½*50*(0.48²-0.4) = 1.8J

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