A light spring with a spring constant of 105.0 N/m rests vertically on the table, as shown in (a) below. A 2.05 g balloon is filled with helium (0°C and 1 atm pressure) to a volume of 4.64 m3 and connected to the spring, causing the spring to stretch, as shown in (b) below. How much does the spring stretch when the system is in equilibrium. (The density of helium is 0.179 kg/m3. The magnitude of the spring force equals kΔx.)

Question

A light spring with a spring constant of 105.0 N/m rests vertically on the table, as shown in (a) below. A 2.05 g balloon is filled with helium (0°C and 1 atm pressure) to a volume of 4.64 m3 and connected to the spring, causing the spring to stretch, as shown in (b) below. How much does the spring stretch when the system is in equilibrium. (The density of helium is 0.179 kg/m3. The magnitude of the spring force equals kΔx.)

Answer 1

mass heliium= volume*density

mass baloon given
weight of air displaced=volume*densityair

spring force=volume*densityair-massHe*g-massbaloon*g

how far it stretches=springforce/k

Answer 2

To determine the amount of spring stretch when the system is in equilibrium, we need to consider the forces acting on the balloon-spring system.

First, let's calculate the mass of the helium inside the balloon. We are given the density of helium as 0.179 kg/m3 and the volume of the balloon as 4.64 m3. We can use the formula:

Mass = Density x Volume

Plugging in the given values:
Mass = 0.179 kg/m3 x 4.64 m3
Mass = 0.829 kg

Next, let's consider the forces acting on the balloon-spring system. The force exerted by the spring is given by Hooke's Law:

Force_spring = Spring Constant x Spring Stretch

We are given the spring constant as 105.0 N/m. Let's assume the spring stretches by Δx meters.

The force exerted by the balloon, which is filled with helium, is given by Archimedes' principle:

Force_balloon = Weight_of_helium
Force_balloon = Mass_of_helium x Gravity

The mass of the helium inside the balloon is 0.829 kg. The weight of the helium can be calculated as:

Weight_of_helium = Mass_of_helium x Gravity
Weight_of_helium = 0.829 kg x 9.8 m/s²
Weight_of_helium = 8.1202 N

Since the system is in equilibrium, the upward force exerted by the balloon must equal the downward force exerted by the spring:

Force_spring = Force_balloon

Plugging in the values we have:

105.0 N/m x Δx = 8.1202 N

Now we can solve for Δx:

Δx = 8.1202 N / 105.0 N/m
Δx = 0.0773 m

Therefore, the spring stretches by 0.0773 meters when the system is in equilibrium.