Question
minimum deceleration required for a 75 kg person to slide down a rope that can sustain a weight of 580 N
Answers
GPT-4o mini
To determine the minimum deceleration required for a 75 kg person to slide down a rope that can sustain a weight of 580 N, we can start by analyzing the forces involved.
1. **Weight of the Person**: The gravitational force acting on the person can be calculated using the following formula:
\[
F = m \cdot g
\]
where:
- \( F \) is the weight (in Newtons),
- \( m \) is the mass (75 kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
Thus, the weight of the person is:
\[
F = 75 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 735.75 \, \text{N}
\]
2. **Maximum Tension in the Rope**: The rope can sustain a maximum weight of 580 N. This means that the rope can only support a downward force of 580 N before it reaches its limit.
3. **Net Force Calculation**: To find the minimum deceleration, we need to compute the net force acting on the person when they are decelerating. The deceleration can create a tension force in addition to their weight.
When the person is sliding down the rope, the net force acting on them can be expressed as:
\[
F_{\text{net}} = T - mg
\]
where:
- \( T \) is the tension in the rope (which can be at maximum 580 N),
- \( mg \) is the weight of the person (735.75 N),
- \( F_{\text{net}} \) is also equal to \( m \cdot a \), where \( a \) is the deceleration. Since it's a downward motion, we’ll take it in the positive direction.
4. **Setting up the equation**: Rearranging gives:
\[
F_{\text{net}} = T - mg = m \cdot (-a)
\]
Where \( -a \) represents deceleration (as it acts in the opposite direction to motion). Substitute the expressions:
\[
580 \, \text{N} - 735.75 \, \text{N} = 75 \, \text{kg} \cdot (-a)
\]
5. **Calculating Deceleration**:
\[
-155.75 \, \text{N} = -75 \, \text{kg} \cdot a
\]
Solving for \( a \):
\[
a = \frac{155.75 \, \text{N}}{75 \, \text{kg}} \approx 2.08 \, \text{m/s}^2
\]
Thus, the minimum deceleration required for a 75 kg person to slide down the rope safely without exceeding the maximum tension in the rope is approximately **2.08 m/s²**.
1. **Weight of the Person**: The gravitational force acting on the person can be calculated using the following formula:
\[
F = m \cdot g
\]
where:
- \( F \) is the weight (in Newtons),
- \( m \) is the mass (75 kg),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)).
Thus, the weight of the person is:
\[
F = 75 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 735.75 \, \text{N}
\]
2. **Maximum Tension in the Rope**: The rope can sustain a maximum weight of 580 N. This means that the rope can only support a downward force of 580 N before it reaches its limit.
3. **Net Force Calculation**: To find the minimum deceleration, we need to compute the net force acting on the person when they are decelerating. The deceleration can create a tension force in addition to their weight.
When the person is sliding down the rope, the net force acting on them can be expressed as:
\[
F_{\text{net}} = T - mg
\]
where:
- \( T \) is the tension in the rope (which can be at maximum 580 N),
- \( mg \) is the weight of the person (735.75 N),
- \( F_{\text{net}} \) is also equal to \( m \cdot a \), where \( a \) is the deceleration. Since it's a downward motion, we’ll take it in the positive direction.
4. **Setting up the equation**: Rearranging gives:
\[
F_{\text{net}} = T - mg = m \cdot (-a)
\]
Where \( -a \) represents deceleration (as it acts in the opposite direction to motion). Substitute the expressions:
\[
580 \, \text{N} - 735.75 \, \text{N} = 75 \, \text{kg} \cdot (-a)
\]
5. **Calculating Deceleration**:
\[
-155.75 \, \text{N} = -75 \, \text{kg} \cdot a
\]
Solving for \( a \):
\[
a = \frac{155.75 \, \text{N}}{75 \, \text{kg}} \approx 2.08 \, \text{m/s}^2
\]
Thus, the minimum deceleration required for a 75 kg person to slide down the rope safely without exceeding the maximum tension in the rope is approximately **2.08 m/s²**.