Answer :
To determine the kinetic energy (KE) of a piece of paper with mass of [tex]\(0.002 \, \text{kg}\)[/tex] falling from a height of [tex]\(8 \, \text{m}\)[/tex], we can start by calculating the potential energy (PE) it has at the initial height using the formula for gravitational potential energy:
[tex]\[ \text{PE} = mgh \][/tex]
where
- [tex]\( m = 0.002 \, \text{kg} \)[/tex] is the mass,
- [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex] is the acceleration due to gravity,
- [tex]\( h = 8 \, \text{m} \)[/tex] is the height.
Plugging in these values:
[tex]\[ \text{PE} = 0.002 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 8 \, \text{m} \][/tex]
[tex]\[ \text{PE} = 0.1568 \, \text{J} \][/tex]
In an ideal, frictionless environment, all this potential energy would convert into kinetic energy as the paper reaches the ground. Hence, the theoretical kinetic energy without any air resistance would be:
[tex]\[ \text{KE}_{\text{ideal}} = 0.1568 \, \text{J} \][/tex]
However, under realistic conditions, the piece of paper experiences air resistance. Air resistance does work against the falling object, and this means that some of the mechanical energy is dissipated as thermal energy due to the friction with air. Consequently, the kinetic energy the piece of paper has as it reaches the ground will be less than the calculated potential energy. Hence:
[tex]\[ \text{KE} < 0.1568 \, \text{J} \][/tex]
Given the options:
A. [tex]\( KE > 0.16 \, \text{J} \)[/tex]
B. [tex]\( KE = 0.16 \, \text{J} \)[/tex]
C. [tex]\( KE = 0 \, \text{J} \)[/tex]
D. [tex]\( KE < 0.16 \, \text{J} \)[/tex]
The correct answer is:
D. [tex]\( \text{KE} < 0.16 \, \text{J} \)[/tex]
[tex]\[ \text{PE} = mgh \][/tex]
where
- [tex]\( m = 0.002 \, \text{kg} \)[/tex] is the mass,
- [tex]\( g = 9.8 \, \text{m/s}^2 \)[/tex] is the acceleration due to gravity,
- [tex]\( h = 8 \, \text{m} \)[/tex] is the height.
Plugging in these values:
[tex]\[ \text{PE} = 0.002 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 8 \, \text{m} \][/tex]
[tex]\[ \text{PE} = 0.1568 \, \text{J} \][/tex]
In an ideal, frictionless environment, all this potential energy would convert into kinetic energy as the paper reaches the ground. Hence, the theoretical kinetic energy without any air resistance would be:
[tex]\[ \text{KE}_{\text{ideal}} = 0.1568 \, \text{J} \][/tex]
However, under realistic conditions, the piece of paper experiences air resistance. Air resistance does work against the falling object, and this means that some of the mechanical energy is dissipated as thermal energy due to the friction with air. Consequently, the kinetic energy the piece of paper has as it reaches the ground will be less than the calculated potential energy. Hence:
[tex]\[ \text{KE} < 0.1568 \, \text{J} \][/tex]
Given the options:
A. [tex]\( KE > 0.16 \, \text{J} \)[/tex]
B. [tex]\( KE = 0.16 \, \text{J} \)[/tex]
C. [tex]\( KE = 0 \, \text{J} \)[/tex]
D. [tex]\( KE < 0.16 \, \text{J} \)[/tex]
The correct answer is:
D. [tex]\( \text{KE} < 0.16 \, \text{J} \)[/tex]
