To determine which object is on the tallest hill, we need to use the formula for gravitational potential energy:
[tex]\[ U = m \cdot g \cdot h \][/tex]
where:
- [tex]\( U \)[/tex] is the potential energy (in joules),
- [tex]\( m \)[/tex] is the mass (in kilograms),
- [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex]),
- [tex]\( h \)[/tex] is the height above the reference point (in meters).
We can rearrange this formula to solve for height [tex]\( h \)[/tex]:
[tex]\[ h = \frac{U}{m \cdot g} \][/tex]
We need to calculate [tex]\( h \)[/tex] for each object.
1. Object W:
[tex]\[ m_W = 50 \, \text{kg} \][/tex]
[tex]\[ U_W = 980 \, \text{J} \][/tex]
[tex]\[ h_W = \frac{980}{50 \cdot 9.8} \approx 2 \, \text{m} \][/tex]
2. Object X:
[tex]\[ m_X = 35 \, \text{kg} \][/tex]
[tex]\[ U_X = 1029 \, \text{J} \][/tex]
[tex]\[ h_X = \frac{1029}{35 \cdot 9.8} \approx 3 \, \text{m} \][/tex]
3. Object Y:
[tex]\[ m_Y = 62 \, \text{kg} \][/tex]
[tex]\[ U_Y = 1519 \, \text{J} \][/tex]
[tex]\[ h_Y = \frac{1519}{62 \cdot 9.8} \approx 2.5 \, \text{m} \][/tex]
4. Object Z:
[tex]\[ m_Z = 24 \, \text{kg} \][/tex]
[tex]\[ U_Z = 1176 \, \text{J} \][/tex]
[tex]\[ h_Z = \frac{1176}{24 \cdot 9.8} \approx 5 \, \text{m} \][/tex]
Examining the calculated heights, we find:
- [tex]\( h_W \approx 2 \, \text{m} \)[/tex]
- [tex]\( h_X \approx 3 \, \text{m} \)[/tex]
- [tex]\( h_Y \approx 2.5 \, \text{m} \)[/tex]
- [tex]\( h_Z \approx 5 \, \text{m} \)[/tex]
Therefore, the object on the tallest hill is Z with a height of approximately 5 meters.
