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To find the tension in the rope when the rock is swinging through its lowest point, follow these steps:
### Step 1: Understand the Energy Conversion
At the highest point (where the rope is horizontal), the rock starts from rest, meaning its initial kinetic energy is zero. As the rock swings down, its potential energy is converted to kinetic energy.
### Step 2: Calculate Potential Energy at the Top
The potential energy (PE) at the top is given by:
[tex]\[ \text{PE}_{\text{top}} = mgh \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the rock: [tex]\( m = 0.500 \)[/tex] kg
- [tex]\( g \)[/tex] is the acceleration due to gravity: [tex]\( g = 9.81 \)[/tex] m/s²
- [tex]\( h \)[/tex] is the height, which in this case is the length of the rope because it's released from the horizontal position: [tex]\( h = 6.00 \)[/tex] m
### Step 3: Set PE Equal to Kinetic Energy at the Lowest Point
At the lowest point, all the potential energy has been converted into kinetic energy (KE):
[tex]\[ \text{PE}_{\text{top}} = \text{KE}_{\text{bottom}} \][/tex]
The kinetic energy at the lowest point is given by:
[tex]\[ \text{KE}_{\text{bottom}} = \frac{1}{2}mv^2 \][/tex]
By conservation of energy:
[tex]\[ mgh = \frac{1}{2}mv^2 \][/tex]
### Step 4: Solve for the Speed [tex]\( v \)[/tex] at the Lowest Point
Cancelling the mass [tex]\( m \)[/tex] from both sides and solving for [tex]\( v \)[/tex]:
[tex]\[ gh = \frac{1}{2}v^2 \][/tex]
[tex]\[ v^2 = 2gh \][/tex]
[tex]\[ v = \sqrt{2gh} \][/tex]
Plugging in the values:
[tex]\[ v = \sqrt{2 \times 9.81 \times 6.00} \][/tex]
[tex]\[ v = 10.85 \text{ m/s} \][/tex]
### Step 5: Calculate the Tension at the Lowest Point
At the lowest point, the tension in the rope has to support both:
1. The centripetal force required to keep the rock moving in a circular path.
2. The weight of the rock.
#### Centripetal Force
The centripetal force [tex]\( F_c \)[/tex] is given by:
[tex]\[ F_c = \frac{mv^2}{r} \][/tex]
where [tex]\( r \)[/tex] is the radius of the circular path (which is the length of the rope):
[tex]\[ r = 6.00 \text{ m} \][/tex]
Plugging in the values:
[tex]\[ F_c = \frac{0.500 \times (10.85)^2}{6.00} \][/tex]
[tex]\[ F_c = 9.81 \text{ N} \][/tex]
#### Weight of the Rock
The weight [tex]\( W \)[/tex] of the rock is given by:
[tex]\[ W = mg \][/tex]
[tex]\[ W = 0.500 \times 9.81 \][/tex]
[tex]\[ W = 4.905 \text{ N} \][/tex]
#### Tension in the Rope
The total tension [tex]\( T \)[/tex] in the rope at the lowest point is the sum of the centripetal force and the weight:
[tex]\[ T = F_c + W \][/tex]
[tex]\[ T = 9.81 + 4.905 \][/tex]
[tex]\[ T = 14.715 \text{ N} \][/tex]
Thus, the tension in the rope when the rock is swinging through its lowest point is [tex]\( 14.715 \)[/tex] N.
### Step 1: Understand the Energy Conversion
At the highest point (where the rope is horizontal), the rock starts from rest, meaning its initial kinetic energy is zero. As the rock swings down, its potential energy is converted to kinetic energy.
### Step 2: Calculate Potential Energy at the Top
The potential energy (PE) at the top is given by:
[tex]\[ \text{PE}_{\text{top}} = mgh \][/tex]
where:
- [tex]\( m \)[/tex] is the mass of the rock: [tex]\( m = 0.500 \)[/tex] kg
- [tex]\( g \)[/tex] is the acceleration due to gravity: [tex]\( g = 9.81 \)[/tex] m/s²
- [tex]\( h \)[/tex] is the height, which in this case is the length of the rope because it's released from the horizontal position: [tex]\( h = 6.00 \)[/tex] m
### Step 3: Set PE Equal to Kinetic Energy at the Lowest Point
At the lowest point, all the potential energy has been converted into kinetic energy (KE):
[tex]\[ \text{PE}_{\text{top}} = \text{KE}_{\text{bottom}} \][/tex]
The kinetic energy at the lowest point is given by:
[tex]\[ \text{KE}_{\text{bottom}} = \frac{1}{2}mv^2 \][/tex]
By conservation of energy:
[tex]\[ mgh = \frac{1}{2}mv^2 \][/tex]
### Step 4: Solve for the Speed [tex]\( v \)[/tex] at the Lowest Point
Cancelling the mass [tex]\( m \)[/tex] from both sides and solving for [tex]\( v \)[/tex]:
[tex]\[ gh = \frac{1}{2}v^2 \][/tex]
[tex]\[ v^2 = 2gh \][/tex]
[tex]\[ v = \sqrt{2gh} \][/tex]
Plugging in the values:
[tex]\[ v = \sqrt{2 \times 9.81 \times 6.00} \][/tex]
[tex]\[ v = 10.85 \text{ m/s} \][/tex]
### Step 5: Calculate the Tension at the Lowest Point
At the lowest point, the tension in the rope has to support both:
1. The centripetal force required to keep the rock moving in a circular path.
2. The weight of the rock.
#### Centripetal Force
The centripetal force [tex]\( F_c \)[/tex] is given by:
[tex]\[ F_c = \frac{mv^2}{r} \][/tex]
where [tex]\( r \)[/tex] is the radius of the circular path (which is the length of the rope):
[tex]\[ r = 6.00 \text{ m} \][/tex]
Plugging in the values:
[tex]\[ F_c = \frac{0.500 \times (10.85)^2}{6.00} \][/tex]
[tex]\[ F_c = 9.81 \text{ N} \][/tex]
#### Weight of the Rock
The weight [tex]\( W \)[/tex] of the rock is given by:
[tex]\[ W = mg \][/tex]
[tex]\[ W = 0.500 \times 9.81 \][/tex]
[tex]\[ W = 4.905 \text{ N} \][/tex]
#### Tension in the Rope
The total tension [tex]\( T \)[/tex] in the rope at the lowest point is the sum of the centripetal force and the weight:
[tex]\[ T = F_c + W \][/tex]
[tex]\[ T = 9.81 + 4.905 \][/tex]
[tex]\[ T = 14.715 \text{ N} \][/tex]
Thus, the tension in the rope when the rock is swinging through its lowest point is [tex]\( 14.715 \)[/tex] N.
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