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princess
princess asked in Science & MathematicsPhysics · 1 decade ago

Physics help please ?????????

A horizontal force is applied to a 2.0 kg box. The box starts from rest, moves a horizontal distance of 12.0 meters, and obtains a velocity of 8.0 m/s. The change in the kinetic energy is

Update:

I got 64 too but the answer key says its 50J.

Any one know why???

5 Answers

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  • shuvo915
    1 decade ago
    Favorite Answer

    Initial kinetic energy K1 = 1/2 x mu² = .5 x 2 x 0 = 0 J

    Final kinetic energy K2 = 1/2 x mv² = .5 x 2 x 8² = 64 J

    So, Change in kinetic energy = 64J

  • Anki
    1 decade ago

    64 N, there is no need for the distance moved by the box! it is simply 1/2 mv^2

  • Hubble
    1 decade ago

    just the answer is 64 J by using the eqn. 1/2mv^2(half mv square)

    haaha one person has written 64 N shame on him

  • sogi
    1 decade ago

    u=0 v = 8m/s m=2kg

    K.E = 1/2 * m * v^2

    change in K.E = 1/2 * m * v^2 - 1/2 * m * u^2

    change in K.E = 1/2 * m * (v^2-u^2)

    put values and you will get answer equal to 64J

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  • ?
    1 decade ago

    kinetic energy = (mxv^2)/2

    so KE= (2x8^2)/2

    KE= 64 J

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