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Answer :
Sure! Let's go through each problem step by step:
Problem 6: Roller Coaster
- Situation: A 500 kg roller coaster starts from rest at the top of an 80.0 m hill. We need to find its speed at the bottom of the hill.
- Concept: Conservation of Energy. The potential energy at the top (due to height) is converted into kinetic energy at the bottom.
1. Potential Energy at the Top:
[tex]\[
\text{PE}_{\text{top}} = m \times g \times h
\][/tex]
where [tex]\( m = 500 \)[/tex] kg, [tex]\( g = 9.81 \)[/tex] m/s², [tex]\( h = 80 \)[/tex] m.
2. Kinetic Energy at the Bottom: Since all potential energy converts into kinetic energy,
[tex]\[
\text{KE}_{\text{bottom}} = \text{PE}_{\text{top}}
\][/tex]
[tex]\[
\frac{1}{2} m v^2 = m \times g \times h
\][/tex]
3. Solve for Speed [tex]\( v \)[/tex]:
[tex]\[
v = \sqrt{2 \times g \times h} = \sqrt{2 \times 9.81 \times 80}
\][/tex]
[tex]\[
v \approx 39.62 \text{ m/s}
\][/tex]
Problem 7: Ball
- Situation: A 2.0 kg ball is tossed straight up with a kinetic energy of 196 joules. We need to find out how high it will go.
- Concept: Conversion of Kinetic Energy to Potential Energy.
1. Kinetic Energy:
[tex]\[
\text{KE} = 196 \text{ Joules}
\][/tex]
2. Potential Energy at Maximum Height: KE converts to PE at the highest point.
[tex]\[
\text{PE}_{\text{top}} = m \times g \times h
\][/tex]
3. Solve for Height [tex]\( h \)[/tex]:
[tex]\[
\text{KE} = m \times g \times h
\][/tex]
[tex]\[
h = \frac{\text{KE}}{m \times g} = \frac{196}{2 \times 9.81}
\][/tex]
[tex]\[
h \approx 9.99 \text{ m}
\][/tex]
Problem 8: Rock
- Situation: A 50 kg rock rolls off the edge of a 30 m cliff. We need to determine how fast it's traveling when it hits the ground.
- Concept: The rock's potential energy converts to kinetic energy as it falls.
1. Initial Potential Energy:
[tex]\[
\text{PE}_{\text{top}} = m \times g \times h
\][/tex]
2. Kinetic Energy at the Ground: Convert initial potential energy to kinetic energy.
[tex]\[
\frac{1}{2} m v^2 = m \times g \times h
\][/tex]
3. Solve for Speed [tex]\( v \)[/tex]:
[tex]\[
v = \sqrt{2 \times g \times h} = \sqrt{2 \times 9.81 \times 30}
\][/tex]
[tex]\[
v \approx 24.26 \text{ m/s}
\][/tex]
These are the speeds and height calculated for each of the scenarios!
Problem 6: Roller Coaster
- Situation: A 500 kg roller coaster starts from rest at the top of an 80.0 m hill. We need to find its speed at the bottom of the hill.
- Concept: Conservation of Energy. The potential energy at the top (due to height) is converted into kinetic energy at the bottom.
1. Potential Energy at the Top:
[tex]\[
\text{PE}_{\text{top}} = m \times g \times h
\][/tex]
where [tex]\( m = 500 \)[/tex] kg, [tex]\( g = 9.81 \)[/tex] m/s², [tex]\( h = 80 \)[/tex] m.
2. Kinetic Energy at the Bottom: Since all potential energy converts into kinetic energy,
[tex]\[
\text{KE}_{\text{bottom}} = \text{PE}_{\text{top}}
\][/tex]
[tex]\[
\frac{1}{2} m v^2 = m \times g \times h
\][/tex]
3. Solve for Speed [tex]\( v \)[/tex]:
[tex]\[
v = \sqrt{2 \times g \times h} = \sqrt{2 \times 9.81 \times 80}
\][/tex]
[tex]\[
v \approx 39.62 \text{ m/s}
\][/tex]
Problem 7: Ball
- Situation: A 2.0 kg ball is tossed straight up with a kinetic energy of 196 joules. We need to find out how high it will go.
- Concept: Conversion of Kinetic Energy to Potential Energy.
1. Kinetic Energy:
[tex]\[
\text{KE} = 196 \text{ Joules}
\][/tex]
2. Potential Energy at Maximum Height: KE converts to PE at the highest point.
[tex]\[
\text{PE}_{\text{top}} = m \times g \times h
\][/tex]
3. Solve for Height [tex]\( h \)[/tex]:
[tex]\[
\text{KE} = m \times g \times h
\][/tex]
[tex]\[
h = \frac{\text{KE}}{m \times g} = \frac{196}{2 \times 9.81}
\][/tex]
[tex]\[
h \approx 9.99 \text{ m}
\][/tex]
Problem 8: Rock
- Situation: A 50 kg rock rolls off the edge of a 30 m cliff. We need to determine how fast it's traveling when it hits the ground.
- Concept: The rock's potential energy converts to kinetic energy as it falls.
1. Initial Potential Energy:
[tex]\[
\text{PE}_{\text{top}} = m \times g \times h
\][/tex]
2. Kinetic Energy at the Ground: Convert initial potential energy to kinetic energy.
[tex]\[
\frac{1}{2} m v^2 = m \times g \times h
\][/tex]
3. Solve for Speed [tex]\( v \)[/tex]:
[tex]\[
v = \sqrt{2 \times g \times h} = \sqrt{2 \times 9.81 \times 30}
\][/tex]
[tex]\[
v \approx 24.26 \text{ m/s}
\][/tex]
These are the speeds and height calculated for each of the scenarios!
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