Answer :
Certainly! Let's find the potential energy of the 25 kg bicycle resting at the top of a 3 m high hill using the formula for gravitational potential energy.
The formula for gravitational potential energy (PE) is:
[tex]\[ PE = m \cdot g \cdot h \][/tex]
Where:
- [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 above the ground (in meters)
Given data:
- Mass ([tex]\( m \)[/tex]) = 25 kg
- Height ([tex]\( h \)[/tex]) = 3 meters
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.8 [tex]\( \text{m/s}^2 \)[/tex]
Now, plug in the values into the formula:
[tex]\[ PE = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3 \, \text{m} \][/tex]
First, multiply the mass by the acceleration due to gravity:
[tex]\[ 25 \times 9.8 = 245 \][/tex]
Next, multiply this result by the height:
[tex]\[ 245 \times 3 = 735 \][/tex]
So, the potential energy of the bicycle is:
[tex]\[ PE = 735 \, \text{Joules (J)} \][/tex]
Therefore, the correct answer is:
[tex]\[ 735 \, \text{J} \][/tex]
The formula for gravitational potential energy (PE) is:
[tex]\[ PE = m \cdot g \cdot h \][/tex]
Where:
- [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 above the ground (in meters)
Given data:
- Mass ([tex]\( m \)[/tex]) = 25 kg
- Height ([tex]\( h \)[/tex]) = 3 meters
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.8 [tex]\( \text{m/s}^2 \)[/tex]
Now, plug in the values into the formula:
[tex]\[ PE = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3 \, \text{m} \][/tex]
First, multiply the mass by the acceleration due to gravity:
[tex]\[ 25 \times 9.8 = 245 \][/tex]
Next, multiply this result by the height:
[tex]\[ 245 \times 3 = 735 \][/tex]
So, the potential energy of the bicycle is:
[tex]\[ PE = 735 \, \text{Joules (J)} \][/tex]
Therefore, the correct answer is:
[tex]\[ 735 \, \text{J} \][/tex]
