My Notes You push a box of mass 24 kg with your car up to an icy hill slope of irregular shape to a height 5.7 m. The box has a speed 12.1 m

Question

My Notes You push a box of mass 24 kg with your car up to an icy hill slope of irregular shape to a height 5.7 m. The box has a speed 12.1 m/s when it starts up the hill, the same time that you brake. It then rises up to the top (with no friction) with a horizontal velocity before immediately falling off a sheer cliff to the ground (with no drag). (a) What is the speed of the box at the top of the hill?

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King 2 months 2021-08-01T00:34:50+00:00 1 Answers 4 views 0

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    2021-08-01T00:36:00+00:00

    Answer:

    The speed of the box at the top of the hill will be 5.693m/s.

    Explanation:

    The kinetic energy of the box at the bottom of the hill is

    K.E = \dfrac{1}{2}mv^2

    putting in m =24kg and  v = 12.1m/s we get

    K.E = \dfrac{1}{2}(24kg)(12.1)^2\\\\K.E = 1756.92J

    Now, the potential energy this box gains as it rises h =5.7m up the hill is

    P.E = mgh

    P.E = (24kg)(10ms/s^2)(5.7m)\\\\P.E = 1368

    Therefore, the energy left E_{left} in the box at the top if the hill will be

    E_{left} =K.E - P.E  = 1756.92J-1368J\\

    \boxed{E_{left} = 388.92J}

    This left-over energy must appear as the kinetic energy of the box at the top of the hill (where else could it go? ); therefore,

    \dfrac{1}{2}mv_t^2= 388.92J

    putting in numbers and solving for v_t we get:

    \boxed{v_t = 5.693m/s.}

    Thus, the speed of the box at the top of the hill is 5.693m/s.

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