The human ciculatory system can be thought of as a closed system of interconnecting pipes through which fluid is continuously circulated by two pumps the two pumps the right and left verticles of the heart, work as simple two-stroke force pumps. The muscles of the heart regulate the force by contracting and relaxing. the contraction (systole) lasts about 0.2s and a complete systole/diastole (contraction/relaxation) cycle lasts about 0.8s. For flood pressures and speeds in the normal range. the volume flow rate of blood through a blood vessel is directly proportional tot he pressure difference over a length of the vessel and to the fourth power of the radius of the vessel. The total mechanical energy per unit volume of blood just as it leaves the heart is `E//V=rhogh+P+rhov^(2)`
Q. What is the gravitational potential energy of `8cm^(3)` of blood in a 1.8 m tall man, in a blood vessel 0.3 m above his heart? (Note: The man's blood pressure is `1.3xx10^(4)N//m^(2)`)

Options

A. `1xx10^(-4)J`
B. `2.5xx10^(-2)J`
C. `301xx10^(3)J`
D. `3xx10^(6)J`

Answer

To find the gravitational potential energy of 8 cm³ of blood in a blood vessel that is 0.3 m above the heart of a 1.8 m tall man, we can use the formula for gravitational potential energy (GPE):

\[ \text{GPE} = \rho \cdot g \cdot h \cdot V \]

Where:
- \( \rho \) = density of blood (approximately \( 1050 \, \text{kg/m}^3 \))
- \( g \) = acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \))
- \( h \) = height above the reference point (0.3 m)
- \( V \) = volume of blood (8 cm³ = \( 8 \times 10^{-6} \, \text{m}^3 \))

### Step 1: Convert Volume from cm³ to m³
The volume given is 8 cm³. To convert this to cubic meters:
\[ V = 8 \, \text{cm}^3 = 8 \times 10^{-6} \, \text{m}^3 \]

### Step 2: Identify the Density of Blood
The density of blood is approximately:
\[ \rho = 1050 \, \text{kg/m}^3 \]

### Step 3: Use the Gravitational Potential Energy Formula
Now we can substitute the values into the GPE formula:
\[ \text{GPE} = \rho \cdot g \cdot h \cdot V \]
Substituting the known values:
\[ \text{GPE} = 1050 \, \text{kg/m}^3 \cdot 9.8 \, \text{m/s}^2 \cdot 0.3 \, \text{m} \cdot 8 \times 10^{-6} \, \text{m}^3 \]

### Step 4: Calculate the GPE
Calculating the above expression step by step:
1. Calculate \( \rho \cdot g \cdot h \):
\[ 1050 \cdot 9.8 \cdot 0.3 = 308.7 \, \text{kg m}^{-2} \text{s}^{-2} \]
2. Now multiply by the volume:
\[ \text{GPE} = 308.7 \cdot 8 \times 10^{-6} = 2.4696 \times 10^{-3} \, \text{J} \]

### Step 5: Final Result
Rounding to two significant figures, we get:
\[ \text{GPE} \approx 2.5 \times 10^{-3} \, \text{J} \]

### Conclusion
The gravitational potential energy of 8 cm³ of blood in a blood vessel 0.3 m above the heart of a 1.8 m tall man is approximately \( 2.5 \times 10^{-3} \, \text{J} \).

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