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The problem provided is about an elevator of mass \( m = 300 \, \text{kg} \) suspended by a steel cable (with negligible mass). We are asked to determine the tension in the cable under two conditions:

1. When the elevator is moving uniformly (constant velocity).
2. When the elevator is moving upwards with a constant acceleration \( a = 0.5 \, \text{m/s}^2 \).

### 1. When the elevator moves uniformly (constant velocity):
In this case, the net force on the elevator is zero because it is not accelerating (Newton's First Law). Thus, the tension \( T \) in the cable equals the gravitational force acting on the elevator.

The gravitational force is given by:
\[
F_g = m \cdot g
\]
where:
- \( m = 300 \, \text{kg} \) is the mass of the elevator,
- \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Thus, the tension when moving at a constant velocity is:
\[
T = m \cdot g = 300 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 2940 \, \text{N}.
\]

### 2. When the elevator is accelerating upwards with \( a = 0.5 \, \text{m/s}^2 \):
In this case, the elevator is accelerating upwards, and the net force is not zero. According to Newton's Second Law, the net force on the elevator is:
\[
F_{\text{net}} = m \cdot a.
\]
The tension in the cable must overcome both the gravitational force and provide the additional force required for the upward acceleration. Therefore, the total tension is:
\[
T = m \cdot (g + a).
\]
Substituting the given values:
\[
T = 300 \, \text{kg} \times (9.8 \, \text{m/s}^2 + 0.5 \, \text{m/s}^2) = 300 \times 10.3 = 3090 \, \text{N}.
\]

### Summary of results:
- **Tension with uniform motion**: \( 2940 \, \text{N} \),
- **Tension with acceleration of \( 0.5 \, \text{m/s}^2 \)**: \( 3090 \, \text{N} \).

Let me know if you'd like further details or explanations! Here are five related questions to deepen understanding:

1. How does the tension change if the elevator is accelerating downwards instead of upwards?
2. What would the tension be if the mass of the cable were not negligible?
3. How would the problem change if the acceleration were larger, say \( 2 \, \text{m/s}^2 \)?
4. What role does the concept of "negligible mass" of the cable play in this problem?
5. How could we model the forces if the elevator were accelerating with variable acceleration?

**Tip**: Whenever you deal with tension problems in elevators, remember to include both gravitational force and any additional forces due to acceleration!

Un ascensore di massa m = 300 kg è appeso a una fune d’acciaio di massa trascurabile. Determinare la tensione della fune quando l’ascensore di muove di moto uniforme e quando sale con un’accelerazione costante a = 0,5 m/s^2.

Calculate the tension in a cable of a 300 kg elevator under uniform motion and when accelerating upward with 0.5 m/s^2. This problem involves key concepts in dynamics and Newton's Laws of Motion, suitable for students in Grades 9-12.

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