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What is the potential energy of a [tex][tex]$25 \, \text{kg}$[/tex][/tex] bicycle resting at the top of a hill [tex][tex]$3 \, \text{m}$[/tex][/tex] high?

(Formula: [tex][tex]$PE = mgh$[/tex][/tex])

A. [tex][tex]$75 \, \text{J}$[/tex][/tex]
B. [tex][tex]$735 \, \text{J}$[/tex][/tex]
C. [tex][tex]$245 \, \text{J}$[/tex][/tex]
D. [tex][tex]$37.8 \, \text{J}$[/tex][/tex]


Sagot :

To find the potential energy of a 25 kg bicycle resting at the top of a hill that is 3 meters high, we use the formula for gravitational potential energy, which is:

[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
- [tex]\( g \)[/tex] is the acceleration due to gravity
- [tex]\( h \)[/tex] is the height from which the object is located

Given:
- [tex]\( m = 25 \)[/tex] kg (the mass of the bicycle)
- [tex]\( g = 9.8 \)[/tex] m/s[tex]\(^2\)[/tex] (the standard acceleration due to gravity)
- [tex]\( h = 3 \)[/tex] m (the height of the hill)

By substituting these values into the formula, we get:

[tex]\[ PE = 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 3 \, \text{m} \][/tex]

First, we calculate the product of the mass and gravity:

[tex]\[ 25 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 245 \, \text{N} \][/tex]

Then, we multiply this result by the height:

[tex]\[ 245 \, \text{N} \times 3 \, \text{m} = 735 \, \text{J} \][/tex]

Therefore, the potential energy of the 25 kg bicycle resting at the top of a 3 meter high hill is:

[tex]\[ 735 \, \text{J} \][/tex]

So, the correct answer is:

[tex]\[ \boxed{735 \, \text{J}} \][/tex]
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