The force, [tex]F[/tex], between two objects is inversely proportional to the square of the distance between them. The distance between two objects is increased by [tex]50 \%[/tex]. Calculate the percent change in the force between the objects.
[tex]\begin{array}{l}
F=\frac{k}{d^2} \\
F=\frac{k}{d(1.5 d)^2} \\
F=\frac{k}{2.25} d^2
\end{array}[/tex]
Answer (1)
The force F is inversely proportional to the square of the distance d : F = d 2 k .
The distance is increased by 50%, so the new distance is d n e w = 1.5 d .
The new force is F n e w = ( 1.5 d ) 2 k = 2.25 d 2 k .
The percentage change in the force is ( F F n e w − 1 ) × 100 = ( 2.25 1 − 1 ) × 100 = − 55.56% .
The force decreases by approximately 55.56% .
Explanation
Initial Setup We are given that the force F between two objects is inversely proportional to the square of the distance d between them. This can be written as F = d 2 k where k is a constant of proportionality.
New Distance The distance between the two objects is increased by 50% . This means the new distance, d n e w , is d n e w = d + 0.5 d = 1.5 d We want to find the percentage change in the force between the objects.
New Force The new force, F n e w , with the new distance d n e w is F n e w = ( 1.5 d ) 2 k = 2.25 d 2 k
Percentage Change Formula To find the percentage change in the force, we use the formula Percentage Change = F F n e w − F × 100 = ( F F n e w − 1 ) × 100 We need to find the ratio F F n e w :
F F n e w = d 2 k 2.25 d 2 k = 2.25 d 2 k × k d 2 = 2.25 1
Calculating the Percentage Change Now, substitute the ratio into the percentage change formula: Percentage Change = ( 2.25 1 − 1 ) × 100 Percentage Change = ( 2.25 1 − 2.25 ) × 100 = 2.25 − 1.25 × 100 = − 2.25 125 = − 9 5 × 100 \text{Percentage Change} = -\frac{500}{9} \approx -55.56%$
Final Answer The percentage change in the force is approximately − 55.56% . This means the force decreases by approximately 55.56% .
Examples
Imagine you are adjusting the distance between a security camera and the area it monitors. If you increase the distance by 50%, the clarity of the image (analogous to the force) will decrease. This calculation helps determine how much the image quality degrades with increased distance, allowing you to make informed decisions about camera placement to maintain adequate surveillance.
