Question

A model airplane with mass 0.5 kg hangs from a rubber band with spring constant 45 N/m. How much is the rubber band stretched when the model airplane hangs motionless?

Answer 1

Multiply 0.5 by 9.8 then divide it by 45. 0.11m

Answer 2

Answer:

0.109 m (10.9 cm)

Explanation:

The weight of the airplane hanging from the rubber band is given by the product between its mass (0.5 kg) and the acceleration due to gravity (g=9.8 m/s^2):

$W=mg=(0.5 kg)(9.8 m/s^2)=4.9 N$

This is the force that stretches the rubber band, according to Hook's law:

$F=kx$

where

k = 45 N/m is the spring constant

x is the stretching of the spring with respect to its equilibrium position

Substituting F = 4.9 N and re-arranging the formula, we get:

$x=\frac{F}{x}=\frac{4.9 N}{45 N/m}=0.109 m=10.9 cm$