Answer :
To find the potential energy of the bicycle at the top of the hill, we'll use the formula for gravitational potential energy:
[tex]\[ PE = m \cdot g \cdot h \][/tex]
where:
- [tex]\( PE \)[/tex] is the potential energy,
- [tex]\( m \)[/tex] is the mass of the object (in kilograms),
- [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex] on Earth),
- [tex]\( h \)[/tex] is the height (in meters).
For this particular problem:
- The mass ([tex]\( m \)[/tex]) of the bicycle is [tex]\( 25 \, \text{kg} \)[/tex].
- The height ([tex]\( h \)[/tex]) of the hill is [tex]\( 3 \, \text{m} \)[/tex].
- The acceleration due to gravity ([tex]\( g \)[/tex]) is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
Now, let's plug these values into the formula:
[tex]\[ PE = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3 \, \text{m} \][/tex]
[tex]\[ PE = 735 \, \text{Joules} \][/tex]
Therefore, the potential energy of the bicycle at the top of the hill is [tex]\( 735 \, \text{J} \)[/tex].
[tex]\[ PE = m \cdot g \cdot h \][/tex]
where:
- [tex]\( PE \)[/tex] is the potential energy,
- [tex]\( m \)[/tex] is the mass of the object (in kilograms),
- [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \)[/tex] on Earth),
- [tex]\( h \)[/tex] is the height (in meters).
For this particular problem:
- The mass ([tex]\( m \)[/tex]) of the bicycle is [tex]\( 25 \, \text{kg} \)[/tex].
- The height ([tex]\( h \)[/tex]) of the hill is [tex]\( 3 \, \text{m} \)[/tex].
- The acceleration due to gravity ([tex]\( g \)[/tex]) is [tex]\( 9.8 \, \text{m/s}^2 \)[/tex].
Now, let's plug these values into the formula:
[tex]\[ PE = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3 \, \text{m} \][/tex]
[tex]\[ PE = 735 \, \text{Joules} \][/tex]
Therefore, the potential energy of the bicycle at the top of the hill is [tex]\( 735 \, \text{J} \)[/tex].
