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    Ldobry7554's Avatar
    Ldobry7554 Posts: 13, Reputation: 1
    New Member

    May 25, 2010, 11:19 AM
    kinetic and potential energy
    A stone weighing 0.2 kilograms is thrown vertically upwards with a velocity of 19.6 meters per second. Neglecting the friction of the air, calculate the kinetic and potential energy it possesses at the end of 1 second and 2 seconds respectively.

    I have this so far: PE=mgh.. therefore it should be 0 because there is no height to account for? Is that correct?

    Also Kinetic energy is 1/2mv^2 which would equal 38.416. However, I don't understand how to account for the 1 and 2 seconds. Any help you could give me?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
    Uber Member

    May 26, 2010, 08:29 AM

    You need to find the height of the stone after 1 second.

    You may use one of the formulae for motion for that: s = ut + 0.5at^2

    where s is the distance the stone covered, u the initial speed, t the time elapsed and a the acceleration the ball experiences, that is the one due to gravity.

    The height after one second is:
    s = (19.6)(1) + 0.5(-9.81)(1)^2 = 14.695 m

    With that height, you can find it's potential energy, and from that you can find the gain in potential energy and hence the kinetic energy using the principle of conservation of energy.

    Do the same for 2 seconds.

    Post your answers! :)
    Ldobry7554's Avatar
    Ldobry7554 Posts: 13, Reputation: 1
    New Member

    May 26, 2010, 11:28 AM

    Thank you!

    So for Height at 2 seconds I got: s=19.6(2) + (-9.81)(2)^2=19.58

    Potential Energy at 1 Seconds= (.2)(9.8)(14.695)=28.8
    Potential Energy at 2 Seconds=(.2)(9.8)(19.58)-38.5

    Kinetic energy is .5(.2)(19.6)^2=38.4

    Is the kinetic energy correct as is or do I still have to account for the seconds?
    Unknown008's Avatar
    Unknown008 Posts: 8,076, Reputation: 723
    Uber Member

    May 27, 2010, 07:49 AM

    Your potential energy at 2 seconds is in fact 38.3768

    [Note that if you use acceleration due to gravity as 9.8 m/s^2 in one formula, use the same value for all the formulae]

    I'll take the first height, using g = 9.8m/s^2 instead of 9.81.

    h1 = (19.6)(1) + 0.5(-9.8)(1)^2 = 14.7 m
    h2 = (19.6)(2) + 0.5(-9.8)(2)^2 = 19.6 m

    Potential energy at (t = 1s) = (0.2)(9.8)(14.7) = 28.812 J (1st answer)
    Potential energy at (t = 2s) = (0.2)(9.8)(19.6) = 38.416 J (2nd answer)

    Now, the kinetic energy initially is = (0.5)(0.2)(19.6)^2 = 38.416 J

    Since no more energy is supplied to the stone, it's total energy will always sum up to 38.416 J

    Total energy = Potential Energy + Kinetic Energy (From principle of conservation of energy)

    But total energy = 38.416 J.

    So, for 1st case;
    38.416 J = 28.812 J + Kinetic energy (3rd answer)

    2nd case;
    38.416 J = 38.416 J + Kinetic energy (4th answer)

    Can you continue now? :)

    Note that since you use g = 9.8 m/s^2, that is you used the value to 2 significant figures, all your answers should also be expressed to 2 significant figures.

    Post your answers.

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