A puck of mass m moving at speed v on a horizontal, frictionless surface is stopped in a distance d because a hockey stick exerts on an oppo

Question

A puck of mass m moving at speed v on a horizontal, frictionless surface is stopped in a distance d because a hockey stick exerts on an opposing force of magnitude F on it.

F = (mv^2) / 2d

If the stopping distance d increases 40%, by what percent does the average force needed to stop the puck change, assuming that m and v are unchanged?

(Fnew – F) / F = ? %

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Khải Quang 2 weeks 2021-08-31T23:10:48+00:00 1 Answers 0 views 0

Answers ( )

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    2021-08-31T23:12:35+00:00

    Answer:

    \% Change = [\frac{1}{2.8} -\frac{1}{1}]*100 = -64.29\%

    Explanation:

    Assuming that m and v are unchanged.

    For this case we have the following formula for the force:

     F = \frac{mv^2}{2d}

    For this case the new force would be given:

     F_{new}= \frac{mv^2}{2*(1.4 d)}

     F_{new}= \frac{mv^2}{2.8 d}

    And for this case we can calculate the % like this:

     \Change = \frac{\frac{mv^2}{2.8d} -\frac{mv^2}{2d}}{\frac{mv^2}{2d}} *100

    And doing the algebra we got:

    \% Change = \frac{\frac{mv^2}{2d}}{\frac{mv^2}{2d}} [\frac{1}{2.8} -\frac{1}{1}]*100

    \% Change = [\frac{1}{2.8} -\frac{1}{1}]*100 = -64.29\%

    So then the force decrease 64.29 percent respect the original force.

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )