A horizontal force is applied to a 4.0 kg box. The box starts from rest, moves a horizontal distance of 10.0 meters, and obtains a velocity

Question

A horizontal force is applied to a 4.0 kg box. The box starts from rest, moves a horizontal distance of 10.0 meters, and obtains a velocity of 7.0 m/s. The change in the kinetic energy is:_____.

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Latifah 18 mins 2021-07-22T14:23:25+00:00 1 Answers 0 views 0

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    2021-07-22T14:24:39+00:00

    Answer:

    98 J

    Explanation:

    Applying,

    Change in kinetic energy = Final kinetic energy- initial kinetic energy

    ΔK.E = mv²/2-mu²/2…………..Equation 1

    Where ΔK.E = Change in kinetic energy, m = mass of the box, u = initial velocity of the box, v = final velocity of the box.

    From the question,

    Given: m = 4.0 kg, u = 0 m/s, v = 7 ,0 m/s

    Substitute these values into equation 1

    ΔK.E = (4(7²)/2)-(4(0²)/2)

    ΔK.E = (2×49)-0

    ΔK.E = 98 J

    Hence the change in kinetic energy 98 J

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