The force, F units, between two objects is inversely proportional to the square of the distance, d units, between them. If the distance decreases by 10%, find the percentage change in the force.

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Worked solution

10

1. Express the inverse-square relationship and set a convenient original distance.

Since FF is inversely proportional to the square of dd,
F=kd2.F=\frac{k}{d^2}.
Let the original distance be dd units, so the original force is
F=kd2.F=\frac{k}{d^2}.

2. Find the new force after the distance decreases by 10%.

A decrease of 10%10\% means the new distance is
0.9d.0.9d.
New force
=k(0.9d)2=k0.81d2=10.81⋅kd2.=\frac{k}{(0.9d)^2}=\frac{k}{0.81d^2}=\frac{1}{0.81}\cdot \frac{k}{d^2}.

3. Calculate the percentage change in the force.

Thus
new forceoriginal force=10.81=1.234567…\frac{\text{new force}}{\text{original force}}=\frac{1}{0.81}=1.234567\ldots
Percentage change
=(1.234567…−1)×100%=23.4567…%.=(1.234567\ldots-1)\times 100\%=23.4567\ldots\%.
So the force increases by 23.5%23.5\% to 3 significant figures.

Answer

The force increases by 23.5%23.5\% to 3 significant figures.

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