Answer :
To find the maximum kinetic energy Dina can reach when she skis to the bottom of the slope, we start by calculating the potential energy at the top of the slope. The equation for potential energy (PE) is:
[tex]\[ PE = m \times g \times h \][/tex]
where:
- \( m \) is the mass (in kilograms),
- \( g \) is the acceleration due to gravity (9.8 m/s²),
- \( h \) is the height (in meters).
Given:
- \( m = 50 \, \text{kg} \),
- \( h = 5 \, \text{m} \),
- \( g = 9.8 \, \text{m/s}^2 \).
Substitute these values into the potential energy formula:
[tex]\[ PE = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 5 \, \text{m} \][/tex]
Calculate the potential energy:
[tex]\[ PE = 50 \times 9.8 \times 5 \][/tex]
[tex]\[ PE = 2450 \, \text{Joules} \][/tex]
Since we are ignoring air resistance and friction, the maximum kinetic energy (KE) Dina can reach at the bottom of the slope will be equal to the potential energy she had at the top. Thus:
[tex]\[ KE_{\text{max}} = 2450 \, \text{Joules} \][/tex]
Therefore, the maximum kinetic energy she can reach is:
[tex]\[ \boxed{2450} \, \text{Joules} \][/tex]
[tex]\[ PE = m \times g \times h \][/tex]
where:
- \( m \) is the mass (in kilograms),
- \( g \) is the acceleration due to gravity (9.8 m/s²),
- \( h \) is the height (in meters).
Given:
- \( m = 50 \, \text{kg} \),
- \( h = 5 \, \text{m} \),
- \( g = 9.8 \, \text{m/s}^2 \).
Substitute these values into the potential energy formula:
[tex]\[ PE = 50 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 5 \, \text{m} \][/tex]
Calculate the potential energy:
[tex]\[ PE = 50 \times 9.8 \times 5 \][/tex]
[tex]\[ PE = 2450 \, \text{Joules} \][/tex]
Since we are ignoring air resistance and friction, the maximum kinetic energy (KE) Dina can reach at the bottom of the slope will be equal to the potential energy she had at the top. Thus:
[tex]\[ KE_{\text{max}} = 2450 \, \text{Joules} \][/tex]
Therefore, the maximum kinetic energy she can reach is:
[tex]\[ \boxed{2450} \, \text{Joules} \][/tex]
