While an elevator of mass 812 kg moves downward, the tension in the supporting cable is a constant 7730 N. Between t= 0 andt3D 4.00 s, the elevator's displacement is 5.00 m downward. What is the elevator's speed at t= 4.00 s? m/s
While an elevator of mass 812 kg moves downward, the tension in the supporting cable is a constant 7730 N. Between t= 0 andt3D 4.00 s, the elevator's displacement is 5.00 m downward. What is the elevator's speed at t= 4.00 s? m/s
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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![**Problem Description:**
While an elevator of mass 812 kg moves downward, the tension in the supporting cable is a constant 7730 N. Between \( t = 0 \) and \( t = 4.00 \) s, the elevator’s displacement is 5.00 m downward. What is the elevator’s speed at \( t = 4.00 \) s?
**Solution:**
To find the elevator’s speed at \( t = 4.00 \) s, we use the following steps:
1. **Calculate the Net Force Acting on the Elevator:**
- Gravitational force (\( F_{\text{gravity}} \)) acting downward:
\[
F_{\text{gravity}} = m \cdot g = 812 \, \text{kg} \times 9.81 \, \text{m/s}^2
\]
- Tension force (\( F_{\text{tension}} \)) acting upward is 7730 N.
- Net force (\( F_{\text{net}} \)) acting on the elevator:
\[
F_{\text{net}} = F_{\text{gravity}} - F_{\text{tension}}
\]
2. **Find the Acceleration of the Elevator:**
Using Newton’s second law:
\[
F_{\text{net}} = m \cdot a
\]
Solve for \( a \) (acceleration):
\[
a = \frac{F_{\text{net}}}{m}
\]
3. **Determine the Speed at \( t = 4.00 \) s:**
Using the kinematic equation:
\[
v = u + a \cdot t
\]
where \( u = 0 \, \text{m/s} \) since the elevator starts from rest, solve for \( v \) (final speed).
**Final Answer:**
After calculating the above steps, fill the value in the box provided in the problem statement.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe9e095e7-b23c-4ebd-957b-6635a59353e5%2F1820d82b-5e73-475e-a03f-7905a8884f55%2Fljkbqje.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
While an elevator of mass 812 kg moves downward, the tension in the supporting cable is a constant 7730 N. Between \( t = 0 \) and \( t = 4.00 \) s, the elevator’s displacement is 5.00 m downward. What is the elevator’s speed at \( t = 4.00 \) s?
**Solution:**
To find the elevator’s speed at \( t = 4.00 \) s, we use the following steps:
1. **Calculate the Net Force Acting on the Elevator:**
- Gravitational force (\( F_{\text{gravity}} \)) acting downward:
\[
F_{\text{gravity}} = m \cdot g = 812 \, \text{kg} \times 9.81 \, \text{m/s}^2
\]
- Tension force (\( F_{\text{tension}} \)) acting upward is 7730 N.
- Net force (\( F_{\text{net}} \)) acting on the elevator:
\[
F_{\text{net}} = F_{\text{gravity}} - F_{\text{tension}}
\]
2. **Find the Acceleration of the Elevator:**
Using Newton’s second law:
\[
F_{\text{net}} = m \cdot a
\]
Solve for \( a \) (acceleration):
\[
a = \frac{F_{\text{net}}}{m}
\]
3. **Determine the Speed at \( t = 4.00 \) s:**
Using the kinematic equation:
\[
v = u + a \cdot t
\]
where \( u = 0 \, \text{m/s} \) since the elevator starts from rest, solve for \( v \) (final speed).
**Final Answer:**
After calculating the above steps, fill the value in the box provided in the problem statement.
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