A man is passing barrels of water to another person from a height of 5 meters. The barrel is attached to a roof that is 10 meters above the ground. The rope holding the barrel is 6 meters long and will break if the tension exceeds 628 N. How many liters of water can the man put in the barrel without the rope breaking (assuming that the barrel is massless)? (The density of water = 1000 kg/m³ and 1 m³ = 1000 L. Density = mass/volume.) Enter a number. 9 m XL 6 m 10 m
A man is passing barrels of water to another person from a height of 5 meters. The barrel is attached to a roof that is 10 meters above the ground. The rope holding the barrel is 6 meters long and will break if the tension exceeds 628 N. How many liters of water can the man put in the barrel without the rope breaking (assuming that the barrel is massless)? (The density of water = 1000 kg/m³ and 1 m³ = 1000 L. Density = mass/volume.) Enter a number. 9 m XL 6 m 10 m
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
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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![**Scenario Description**
A man is passing barrels of water to another person from a height of 5 meters. The barrel is attached to a roof that is 10 meters above the ground. The rope holding the barrel is 6 meters long and will break if the tension exceeds 628 Newtons. The question asks how many liters of water can the man put in the barrel without the rope breaking, assuming the barrel is massless.
**Given Data**
- Height from which the man is passing barrels: 5 m
- Height of the roof: 10 m
- Length of the rope: 6 m
- Maximum tension in the rope before breaking: 628 N
- Density of water: 1000 kg/m³
- Volume conversion: 1 m³ = 1000 L
**Diagram Explanation**
The diagram shows:
- A barrel initially at a height with a downward arrow indicating its movement to another position.
- A left vertical measurement labeled 9 m.
- Vertical measurement of the rope length, marked as 6 m, running from the top roof (10 m above the ground) to the suspended barrel.
**Instructions**
To solve the problem, use the following approach:
1. Understand the relation between tension, mass, and gravitational force.
2. The formula for tension is \( T = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximated as 9.8 m/s²).
3. Calculate the maximum permissible mass that can be supported by the rope without breaking it.
4. Convert this mass to volume using the density of water.
5. Convert this volume to liters using the conversion factor.
**Problem Solution Steps**
1. Calculate the maximum mass the rope can support without breaking:
\[
T = mg \Rightarrow m = \frac{T}{g} = \frac{628 \, \text{N}}{9.8 \, \text{m/s}^2}
\]
2. Convert mass to volume using the density of water:
\[
\text{Volume} = \frac{m}{\text{Density}} = \frac{m}{1000 \, \text{kg/m}^3}
\]
3. Convert cubic meters to liters:
\[
\text{Liters} = \text{Volume} \times 1000 \, \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fe7370c0b-05cc-48e6-9b0a-29a272be895a%2F34b8c311-c8cb-4a98-af8e-3b733890a2d9%2F2ih1ba_processed.png&w=3840&q=75)
Transcribed Image Text:**Scenario Description**
A man is passing barrels of water to another person from a height of 5 meters. The barrel is attached to a roof that is 10 meters above the ground. The rope holding the barrel is 6 meters long and will break if the tension exceeds 628 Newtons. The question asks how many liters of water can the man put in the barrel without the rope breaking, assuming the barrel is massless.
**Given Data**
- Height from which the man is passing barrels: 5 m
- Height of the roof: 10 m
- Length of the rope: 6 m
- Maximum tension in the rope before breaking: 628 N
- Density of water: 1000 kg/m³
- Volume conversion: 1 m³ = 1000 L
**Diagram Explanation**
The diagram shows:
- A barrel initially at a height with a downward arrow indicating its movement to another position.
- A left vertical measurement labeled 9 m.
- Vertical measurement of the rope length, marked as 6 m, running from the top roof (10 m above the ground) to the suspended barrel.
**Instructions**
To solve the problem, use the following approach:
1. Understand the relation between tension, mass, and gravitational force.
2. The formula for tension is \( T = mg \), where \( m \) is the mass and \( g \) is the acceleration due to gravity (approximated as 9.8 m/s²).
3. Calculate the maximum permissible mass that can be supported by the rope without breaking it.
4. Convert this mass to volume using the density of water.
5. Convert this volume to liters using the conversion factor.
**Problem Solution Steps**
1. Calculate the maximum mass the rope can support without breaking:
\[
T = mg \Rightarrow m = \frac{T}{g} = \frac{628 \, \text{N}}{9.8 \, \text{m/s}^2}
\]
2. Convert mass to volume using the density of water:
\[
\text{Volume} = \frac{m}{\text{Density}} = \frac{m}{1000 \, \text{kg/m}^3}
\]
3. Convert cubic meters to liters:
\[
\text{Liters} = \text{Volume} \times 1000 \, \
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