Answer :
To solve this problem, we're given the mass of the cylinders and the height. We need to calculate the gravitational potential energy for each mass. The formula for gravitational potential energy [tex]\( P.E \)[/tex] is given by:
[tex]\[ P.E = m \cdot g \cdot h \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass in kilograms (kg)
- [tex]\( g \)[/tex] is the acceleration due to gravity, which is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]
- [tex]\( h \)[/tex] is the height in meters (m)
### Step-by-Step Solution:
1. Identify the values given:
- [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- [tex]\( h = 500 \, \text{meters} \)[/tex]
- Masses [tex]\( m = 3 \, \text{kg} \)[/tex], [tex]\( 6 \, \text{kg} \)[/tex], and [tex]\( 9 \, \text{kg} \)[/tex]
2. Calculate the gravitational potential energy for each mass:
- For [tex]\( m = 3 \, \text{kg} \)[/tex]:
[tex]\[ P.E_{\text{3kg}} = 3 \cdot 9.8 \cdot 500 \][/tex]
[tex]\[ P.E_{\text{3kg}} = 14700 \, \text{J} \][/tex]
To convert joules (J) to kilojoules (kJ), we divide by 1000:
[tex]\[ P.E_{\text{3kg}} = \frac{14700}{1000} = 14.7 \, \text{kJ} \][/tex]
- For [tex]\( m = 6 \, \text{kg} \)[/tex]:
[tex]\[ P.E_{\text{6kg}} = 6 \cdot 9.8 \cdot 500 \][/tex]
[tex]\[ P.E_{\text{6kg}} = 29400 \, \text{J} \][/tex]
To convert joules (J) to kilojoules (kJ), we divide by 1000:
[tex]\[ P.E_{\text{6kg}} = \frac{29400}{1000} = 29.4 \, \text{kJ} \][/tex]
- For [tex]\( m = 9 \, \text{kg} \)[/tex]:
[tex]\[ P.E_{\text{9kg}} = 9 \cdot 9.8 \cdot 500 \][/tex]
[tex]\[ P.E_{\text{9kg}} = 44100 \, \text{J} \][/tex]
To convert joules (J) to kilojoules (kJ), we divide by 1000:
[tex]\[ P.E_{\text{9kg}} = \frac{44100}{1000} = 44.1 \, \text{kJ} \][/tex]
### Summary:
- Gravitational potential energy for [tex]\( 3 \, \text{kg} \)[/tex] mass is [tex]\( 14.7 \, \text{kJ} \)[/tex].
- Gravitational potential energy for [tex]\( 6 \, \text{kg} \)[/tex] mass is [tex]\( 29.4 \, \text{kJ} \)[/tex].
- Gravitational potential energy for [tex]\( 9 \, \text{kg} \)[/tex] mass is [tex]\( 44.1 \, \text{kJ} \)[/tex].
Thus:
- 3 kg: [tex]\( 14.7 \, \text{kJ} \)[/tex]
- 6 kg: [tex]\( 29.4 \, \text{kJ} \)[/tex]
- 9 kg: [tex]\( 44.1 \, \text{kJ} \)[/tex]
[tex]\[ P.E = m \cdot g \cdot h \][/tex]
Where:
- [tex]\( m \)[/tex] is the mass in kilograms (kg)
- [tex]\( g \)[/tex] is the acceleration due to gravity, which is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]
- [tex]\( h \)[/tex] is the height in meters (m)
### Step-by-Step Solution:
1. Identify the values given:
- [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex]
- [tex]\( h = 500 \, \text{meters} \)[/tex]
- Masses [tex]\( m = 3 \, \text{kg} \)[/tex], [tex]\( 6 \, \text{kg} \)[/tex], and [tex]\( 9 \, \text{kg} \)[/tex]
2. Calculate the gravitational potential energy for each mass:
- For [tex]\( m = 3 \, \text{kg} \)[/tex]:
[tex]\[ P.E_{\text{3kg}} = 3 \cdot 9.8 \cdot 500 \][/tex]
[tex]\[ P.E_{\text{3kg}} = 14700 \, \text{J} \][/tex]
To convert joules (J) to kilojoules (kJ), we divide by 1000:
[tex]\[ P.E_{\text{3kg}} = \frac{14700}{1000} = 14.7 \, \text{kJ} \][/tex]
- For [tex]\( m = 6 \, \text{kg} \)[/tex]:
[tex]\[ P.E_{\text{6kg}} = 6 \cdot 9.8 \cdot 500 \][/tex]
[tex]\[ P.E_{\text{6kg}} = 29400 \, \text{J} \][/tex]
To convert joules (J) to kilojoules (kJ), we divide by 1000:
[tex]\[ P.E_{\text{6kg}} = \frac{29400}{1000} = 29.4 \, \text{kJ} \][/tex]
- For [tex]\( m = 9 \, \text{kg} \)[/tex]:
[tex]\[ P.E_{\text{9kg}} = 9 \cdot 9.8 \cdot 500 \][/tex]
[tex]\[ P.E_{\text{9kg}} = 44100 \, \text{J} \][/tex]
To convert joules (J) to kilojoules (kJ), we divide by 1000:
[tex]\[ P.E_{\text{9kg}} = \frac{44100}{1000} = 44.1 \, \text{kJ} \][/tex]
### Summary:
- Gravitational potential energy for [tex]\( 3 \, \text{kg} \)[/tex] mass is [tex]\( 14.7 \, \text{kJ} \)[/tex].
- Gravitational potential energy for [tex]\( 6 \, \text{kg} \)[/tex] mass is [tex]\( 29.4 \, \text{kJ} \)[/tex].
- Gravitational potential energy for [tex]\( 9 \, \text{kg} \)[/tex] mass is [tex]\( 44.1 \, \text{kJ} \)[/tex].
Thus:
- 3 kg: [tex]\( 14.7 \, \text{kJ} \)[/tex]
- 6 kg: [tex]\( 29.4 \, \text{kJ} \)[/tex]
- 9 kg: [tex]\( 44.1 \, \text{kJ} \)[/tex]