Question

A consumer upset with the latest trend of postal rate increases has decided to try to send letters by balloon even though they may not reach their intended destination. A 6.00 times 10^4 cm^3 gas-filled balloon will provide enough lift for a 40.0 g package to be accelerated upward at a rate of 2.65 m/s^2. For these circumstances, calculate the density of the gas the consumer fills the balloon with. The acceleration due to gravity is g = 9.81 m/s^2 and the density of air is rho_air r = 1.16 kg/m^3. Neglect the mass of the balloon material and the volume of the package.

Answer 1

We know that the formula for the Buoyancy force is given by

Fb =ma+mg

pgV =ma+mg

pgV =m(a+g)

Then mass (m) =pgV/(a+g) =(1.16kg/m3)(9.81m/s2)(6*10^-2m³)/(2.65m/s2+9.81m/s2)=0.054797kg

Now the mass of the air balloon is given by

m' =m -mp =0.054797kg -0.04kg=0.014797kg

Then the density of the gas consumer is (p) =mass/volume =0.014797kg/6*10^-2m³=0.2466kg/m³