Answer :
Certainly! Let's go through each part of the problem step-by-step:
### Part (a): Work Done to Fill the Tank When Upright
1. Tank Dimensions (Upright Position)
- Height: [tex]\( h_u = 2 \, \text{m} \)[/tex]
- Base Area: [tex]\( A_u = 1 \, \text{m}^2 \)[/tex] (since the base is a [tex]\(1 \, \text{m} \times 1 \, \text{m}\)[/tex] square)
2. Calculating Work Done
- To find the work done to pump water to a height [tex]\(h\)[/tex], consider a small slice of water with thickness [tex]\(dh\)[/tex] at height [tex]\(h\)[/tex].
- The volume of this slice is [tex]\( dV = A_u \, dh \)[/tex].
- The mass of this slice is [tex]\( dm = \rho \, dV = \rho \, A_u \, dh \)[/tex].
- The gravitational force on this slice is [tex]\( dF = dm \, g = \rho \, A_u \, g \, dh \)[/tex].
- The work done to lift this slice to height [tex]\(h\)[/tex] is [tex]\( dW = dF \cdot h = \rho \, A_u \, g \, h \, dh \)[/tex].
3. Integrating to Find Total Work
- Integrate from [tex]\( h = 0 \)[/tex] to [tex]\( h = h_u \)[/tex]:
[tex]\[ W_u = \int_0^{h_u} \rho \, A_u \, g \, h \, dh = \rho \, A_u \, g \int_0^{h_u} h \, dh = \rho \, A_u \, g \left[\frac{h^2}{2}\right]_0^{h_u} \][/tex]
- Substituting the given values:
[tex]\[ W_u = 1000 \, \text{kg/m}^3 \times 1 \, \text{m}^2 \times 9.8 \, \text{m/s}^2 \times \frac{2^2}{2} = 19600 \, \text{J} \][/tex]
Hence, the work done to fill the tank when it is upright is 19600 J.
### Part (b): Work Done to Fill the Tank When Lying on the Side
1. Tank Dimensions (Side Position)
- Height: [tex]\( h_s = 1 \, \text{m} \)[/tex]
- Base Area: [tex]\( A_s = 2 \, \text{m}^2 \)[/tex] (since the base is a [tex]\(2 \, \text{m} \times 1 \, \text{m}\)[/tex] rectangle)
2. Calculating Work Done
- Following a similar integration approach:
[tex]\[ W_s = \int_0^{h_s} \rho \, A_s \, g \, h \, dh = \rho \, A_s \, g \int_0^{h_s} h \, dh = \rho \, A_s \, g \left[\frac{h^2}{2}\right]_0^{h_s} \][/tex]
- Substituting the given values:
[tex]\[ W_s = 1000 \, \text{kg/m}^3 \times 2 \, \text{m}^2 \times 9.8 \, \text{m/s}^2 \times \frac{1^2}{2} = 9800 \, \text{J} \][/tex]
Hence, the work done to fill the tank when it is lying on its side is 9800 J.
### Part (c): Work Done to Fill the Tank by Pouring Into the Top
1. Work Done (Upright Position)
- The mass of water to fill the tank is [tex]\( m = \rho \, \text{Volume} = 1000 \, \text{kg/m}^3 \times 2 \, \text{m}^3 = 2000 \, \text{kg} \)[/tex].
- The center of mass will be at the mid-height of the tank, i.e., [tex]\( 1 \, \text{m} \)[/tex] from the bottom.
- The work done to raise the center of mass to this height:
[tex]\[ W_{u, \text{top}} = m \times g \times \left(\frac{h_u}{2}\right) = 2000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1 \, \text{m} = 19600 \, \text{J} \][/tex]
2. Work Done (Side Position)
- Using the same mass of water, [tex]\( m = 2000 \, \text{kg} \)[/tex].
- The center of mass in this case is at [tex]\( \frac{h_s}{2} = 0.5 \, \text{m} \)[/tex] from the bottom.
[tex]\[ W_{s, \text{top}} = m \times g \times \left(\frac{h_s}{2}\right) = 2000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.5 \, \text{m} = 9800 \, \text{J} \][/tex]
Hence, the work done to fill the tank by pouring into the top is 19600 J for the upright position and 9800 J for the side position.
### Comparison and Explanation
- Difference in Work Required:
- Filling from the bottom up requires more work in both positions as compared to filling from the top. This is because when we fill from the bottom, we continuously work against gravity to lift each slice of water. The further we go up, the more potential energy we have to overcome.
- When pouring from the top, we only need to lift the entire mass to the height of the center of mass once. This simplifies the process since all water is considered being lifted uniformly.
- Why are They Different:
- The primary difference stems from the distribution of the effort over the height in the bottom-up filling method versus lifting once to the center of mass in the top-filling method.
### Part (a): Work Done to Fill the Tank When Upright
1. Tank Dimensions (Upright Position)
- Height: [tex]\( h_u = 2 \, \text{m} \)[/tex]
- Base Area: [tex]\( A_u = 1 \, \text{m}^2 \)[/tex] (since the base is a [tex]\(1 \, \text{m} \times 1 \, \text{m}\)[/tex] square)
2. Calculating Work Done
- To find the work done to pump water to a height [tex]\(h\)[/tex], consider a small slice of water with thickness [tex]\(dh\)[/tex] at height [tex]\(h\)[/tex].
- The volume of this slice is [tex]\( dV = A_u \, dh \)[/tex].
- The mass of this slice is [tex]\( dm = \rho \, dV = \rho \, A_u \, dh \)[/tex].
- The gravitational force on this slice is [tex]\( dF = dm \, g = \rho \, A_u \, g \, dh \)[/tex].
- The work done to lift this slice to height [tex]\(h\)[/tex] is [tex]\( dW = dF \cdot h = \rho \, A_u \, g \, h \, dh \)[/tex].
3. Integrating to Find Total Work
- Integrate from [tex]\( h = 0 \)[/tex] to [tex]\( h = h_u \)[/tex]:
[tex]\[ W_u = \int_0^{h_u} \rho \, A_u \, g \, h \, dh = \rho \, A_u \, g \int_0^{h_u} h \, dh = \rho \, A_u \, g \left[\frac{h^2}{2}\right]_0^{h_u} \][/tex]
- Substituting the given values:
[tex]\[ W_u = 1000 \, \text{kg/m}^3 \times 1 \, \text{m}^2 \times 9.8 \, \text{m/s}^2 \times \frac{2^2}{2} = 19600 \, \text{J} \][/tex]
Hence, the work done to fill the tank when it is upright is 19600 J.
### Part (b): Work Done to Fill the Tank When Lying on the Side
1. Tank Dimensions (Side Position)
- Height: [tex]\( h_s = 1 \, \text{m} \)[/tex]
- Base Area: [tex]\( A_s = 2 \, \text{m}^2 \)[/tex] (since the base is a [tex]\(2 \, \text{m} \times 1 \, \text{m}\)[/tex] rectangle)
2. Calculating Work Done
- Following a similar integration approach:
[tex]\[ W_s = \int_0^{h_s} \rho \, A_s \, g \, h \, dh = \rho \, A_s \, g \int_0^{h_s} h \, dh = \rho \, A_s \, g \left[\frac{h^2}{2}\right]_0^{h_s} \][/tex]
- Substituting the given values:
[tex]\[ W_s = 1000 \, \text{kg/m}^3 \times 2 \, \text{m}^2 \times 9.8 \, \text{m/s}^2 \times \frac{1^2}{2} = 9800 \, \text{J} \][/tex]
Hence, the work done to fill the tank when it is lying on its side is 9800 J.
### Part (c): Work Done to Fill the Tank by Pouring Into the Top
1. Work Done (Upright Position)
- The mass of water to fill the tank is [tex]\( m = \rho \, \text{Volume} = 1000 \, \text{kg/m}^3 \times 2 \, \text{m}^3 = 2000 \, \text{kg} \)[/tex].
- The center of mass will be at the mid-height of the tank, i.e., [tex]\( 1 \, \text{m} \)[/tex] from the bottom.
- The work done to raise the center of mass to this height:
[tex]\[ W_{u, \text{top}} = m \times g \times \left(\frac{h_u}{2}\right) = 2000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1 \, \text{m} = 19600 \, \text{J} \][/tex]
2. Work Done (Side Position)
- Using the same mass of water, [tex]\( m = 2000 \, \text{kg} \)[/tex].
- The center of mass in this case is at [tex]\( \frac{h_s}{2} = 0.5 \, \text{m} \)[/tex] from the bottom.
[tex]\[ W_{s, \text{top}} = m \times g \times \left(\frac{h_s}{2}\right) = 2000 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 0.5 \, \text{m} = 9800 \, \text{J} \][/tex]
Hence, the work done to fill the tank by pouring into the top is 19600 J for the upright position and 9800 J for the side position.
### Comparison and Explanation
- Difference in Work Required:
- Filling from the bottom up requires more work in both positions as compared to filling from the top. This is because when we fill from the bottom, we continuously work against gravity to lift each slice of water. The further we go up, the more potential energy we have to overcome.
- When pouring from the top, we only need to lift the entire mass to the height of the center of mass once. This simplifies the process since all water is considered being lifted uniformly.
- Why are They Different:
- The primary difference stems from the distribution of the effort over the height in the bottom-up filling method versus lifting once to the center of mass in the top-filling method.
