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Sagot :
To determine the tension in the string for a body of mass \(2 \text{ kg}\) moving with constant speed in a horizontal circle of radius \(200 \text{ cm}\) attached to a string of \(4 \text{ m}\), we can follow the steps below.
### Given Data:
- Mass of the body, \( m = 2 \text{ kg} \)
- Radius of the circular path, \( r = 200 \text{ cm} = 2 \text{ m} \) (since \(1 \text{ m} = 100 \text{ cm}\))
- Length of the string, \( L = 4 \text{ m} \)
- Acceleration due to gravity, \( g = 10 \text{ m/s}^2 \)
### Understanding the Situation:
In this scenario, a body is moving in a horizontal circle. The centripetal force required to keep the body in circular motion is provided by the tension in the string. The tension in the string needs to counteract the gravitational force as well as provide the necessary centripetal force for circular motion.
### Calculating the Tension:
Since the body is moving in a horizontal circle, we need to first consider the forces acting on the body:
1. Gravitational Force (Weight): \( F_g = m \cdot g \)
2. Centripetal Force (provided by tension in the string): \( F_c = \frac{m v^2}{r} \)
However, in the scenario where the body is in horizontal circular motion, the centripetal force does not need explicit calculation of velocity \(v\) since it is implied that gravitational force will be balanced by the vertical component of the tension (for simplicity, we consider the vertical component).
Since we're given values only relevant to the gravitational force and mass, we focus on calculating using these parameters.
### Step-by-Step Solution:
1. Calculate the gravitational force:
[tex]\[ F_g = m \cdot g = 2 \text{ kg} \cdot 10 \text{ m/s}^2 = 20 \text{ N} \][/tex]
2. Determining the Tension:
The tension \( T \) in the string must balance out the gravitational force vertically, which results in:
[tex]\[ T = F_g = 20 \text{ N} \][/tex]
The vertical component of the tension is equal to the weight of the mass. Therefore, the tension in the string is \( 20 \text{ N} \).
### Conclusion:
The tension in the string when a body of mass [tex]\(2 \text{ kg}\)[/tex] is moving in a horizontal circle of radius [tex]\(200 \text{ cm}\)[/tex] with the help of a string of length [tex]\(4 \text{ m}\)[/tex] is [tex]\( 20 \text{ N} \)[/tex].
### Given Data:
- Mass of the body, \( m = 2 \text{ kg} \)
- Radius of the circular path, \( r = 200 \text{ cm} = 2 \text{ m} \) (since \(1 \text{ m} = 100 \text{ cm}\))
- Length of the string, \( L = 4 \text{ m} \)
- Acceleration due to gravity, \( g = 10 \text{ m/s}^2 \)
### Understanding the Situation:
In this scenario, a body is moving in a horizontal circle. The centripetal force required to keep the body in circular motion is provided by the tension in the string. The tension in the string needs to counteract the gravitational force as well as provide the necessary centripetal force for circular motion.
### Calculating the Tension:
Since the body is moving in a horizontal circle, we need to first consider the forces acting on the body:
1. Gravitational Force (Weight): \( F_g = m \cdot g \)
2. Centripetal Force (provided by tension in the string): \( F_c = \frac{m v^2}{r} \)
However, in the scenario where the body is in horizontal circular motion, the centripetal force does not need explicit calculation of velocity \(v\) since it is implied that gravitational force will be balanced by the vertical component of the tension (for simplicity, we consider the vertical component).
Since we're given values only relevant to the gravitational force and mass, we focus on calculating using these parameters.
### Step-by-Step Solution:
1. Calculate the gravitational force:
[tex]\[ F_g = m \cdot g = 2 \text{ kg} \cdot 10 \text{ m/s}^2 = 20 \text{ N} \][/tex]
2. Determining the Tension:
The tension \( T \) in the string must balance out the gravitational force vertically, which results in:
[tex]\[ T = F_g = 20 \text{ N} \][/tex]
The vertical component of the tension is equal to the weight of the mass. Therefore, the tension in the string is \( 20 \text{ N} \).
### Conclusion:
The tension in the string when a body of mass [tex]\(2 \text{ kg}\)[/tex] is moving in a horizontal circle of radius [tex]\(200 \text{ cm}\)[/tex] with the help of a string of length [tex]\(4 \text{ m}\)[/tex] is [tex]\( 20 \text{ N} \)[/tex].
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