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10 December, 02:42

After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. When the box reaches position x2 (where x2>x1), how much work Wp has the person done on the box

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  1. 10 December, 05:09
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    0.5m (v1) ²

    Explanation:

    We know that work done on an object is equal to the change in the kinetic energy of that object. Hence,

    W. D = ΔK. E.

    ΔK. E. = K. E. (final) - K. E. (initial)

    Initial Kinetic energy, K. E. (initial), is:

    K. E. (initial) = 0 (the box was at West)

    Final K. E., K. E. (final) is:

    K. E. (final) = 0.5*m * (v1) ²

    Hence, work done on the box will be:

    W. D. = 0.5m (v1) ² - 0

    W. D. = 0.5m (v1) ²
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