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 = ? %

Answer 1

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.