A light spring of constant k = 95.0 N/m is attached vertically to a table (figure (a)). A 2.70-g balloon is filled with helium (density = 0.179 kg/m3) to a volume of 4.00 m3 and is then connected to the spring, causing the spring to stretch as shown in figure (b). Determine the extension distance L when the balloon is in equilibrium. (The density of air is 1.29 kg/m3.)
m
mass balloon = .0027 kg
buoyancy net up force = g [ 4 (1.29-.179) -.0027 ]
= g (4.44)
= 43.6 N using g = 9.81 m/s^2
F = k x
43.6 = 95 x
x = .459 N
To determine the extension distance L when the balloon is in equilibrium, we need to consider the forces acting on the system.
First, let's consider the weight of the balloon. The weight is given by the formula:
Weight = Mass * Acceleration due to gravity
The mass of the balloon can be found using the density formula:
Mass = Density * Volume
In this case, the density of helium is given as 0.179 kg/m^3 and the volume is given as 4.00 m^3. Therefore, the mass of the balloon is:
Mass = 0.179 kg/m^3 * 4.00 m^3 = 0.716 kg
The weight of the balloon can be calculated using the acceleration due to gravity, which is approximately 9.8 m/s^2:
Weight = 0.716 kg * 9.8 m/s^2 = 7.01 N
Since the balloon is in equilibrium, the weight of the balloon is balanced by the force exerted by the spring. According to Hooke's Law, the force exerted by a spring is given by:
Force = Spring Constant * Extension
In this case, the spring constant is given as 95.0 N/m. Therefore, the force exerted by the spring is:
Force = 95.0 N/m * Extension
Since the weight and the spring force are equal in equilibrium, we have:
7.01 N = 95.0 N/m * Extension
Now, we can solve for the extension distance L:
Extension = 7.01 N / (95.0 N/m) = 0.0737 m
Therefore, the extension distance L when the balloon is in equilibrium is approximately 0.0737 m.