(b) The coefficient of kinetic friction between a 40 kg crate and the ware- house floor is 70% of the corresponding coefficient of static friction. The crate falls off a forklift that is moving at 3 m/s and then slides along the warehouse floor for a distance of 2.5 m before coming to rest. What is the coefficient of static friction between the crate and the floor?
(b) The coefficient of kinetic friction between a 40 kg crate and the ware- house floor is 70% of the corresponding coefficient of static friction. The crate falls off a forklift that is moving at 3 m/s and then slides along the warehouse floor for a distance of 2.5 m before coming to rest. What is the coefficient of static friction between the crate and the floor?
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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![### Problem Statement:
**Scenario:**
A 40 kg crate falls off a forklift moving at 3 m/s and slides along the warehouse floor, coming to rest after covering a distance of 2.5 m.
**Given Data:**
- Mass of the crate (\(m\)): 40 kg
- Initial velocity (\(v_i\)): 3 m/s
- Distance traveled before stopping (\(d\)): 2.5 m
- Coefficient of kinetic friction (\(\mu_k\)): 70% of the coefficient of static friction (\(\mu_s\))
**Question:**
What is the coefficient of static friction (\(\mu_s\)) between the crate and the warehouse floor?
### Explanation and Solution:
1. **Kinetic Friction and Stopping Distance:**
When the crate falls and slides to a stop, kinetic friction is the force that brings it to rest. The work done by this frictional force over the distance of 2.5 m is equal to the kinetic energy the crate had initially.
2. **Equations Involved:**
- Kinetic energy (\(KE\)): \[ KE = \frac{1}{2} m v_i^2 \]
- Work done by friction (\(W\)): \[ W = F_k \cdot d \]
- Force of kinetic friction (\(F_k\)): \[ F_k = \mu_k \cdot N \]
where \(N\) is the normal force, \(N = m \cdot g\).
3. **Steps to Find \(\mu_s\):**
1. Calculate the initial kinetic energy (\(KE\)).
2. Set up the work-energy equation: \( KE = F_k \cdot d \).
3. Rearrange to find \(F_k \).
4. Calculate \(\mu_k\).
5. Use the given relationship \(\mu_k = 0.7 \mu_s\) to solve for \(\mu_s\).
**Detailed Solution:**
1. **Calculate initial kinetic energy:**
\[ KE = \frac{1}{2} \cdot 40 \, \text{kg} \cdot (3 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \cdot 40 \cdot](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5623d6a3-334a-4b1a-9cf6-41cb3cd5d075%2F665f5537-0f89-41af-a27a-87cb06409e71%2Fjonm2ph_processed.png&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
**Scenario:**
A 40 kg crate falls off a forklift moving at 3 m/s and slides along the warehouse floor, coming to rest after covering a distance of 2.5 m.
**Given Data:**
- Mass of the crate (\(m\)): 40 kg
- Initial velocity (\(v_i\)): 3 m/s
- Distance traveled before stopping (\(d\)): 2.5 m
- Coefficient of kinetic friction (\(\mu_k\)): 70% of the coefficient of static friction (\(\mu_s\))
**Question:**
What is the coefficient of static friction (\(\mu_s\)) between the crate and the warehouse floor?
### Explanation and Solution:
1. **Kinetic Friction and Stopping Distance:**
When the crate falls and slides to a stop, kinetic friction is the force that brings it to rest. The work done by this frictional force over the distance of 2.5 m is equal to the kinetic energy the crate had initially.
2. **Equations Involved:**
- Kinetic energy (\(KE\)): \[ KE = \frac{1}{2} m v_i^2 \]
- Work done by friction (\(W\)): \[ W = F_k \cdot d \]
- Force of kinetic friction (\(F_k\)): \[ F_k = \mu_k \cdot N \]
where \(N\) is the normal force, \(N = m \cdot g\).
3. **Steps to Find \(\mu_s\):**
1. Calculate the initial kinetic energy (\(KE\)).
2. Set up the work-energy equation: \( KE = F_k \cdot d \).
3. Rearrange to find \(F_k \).
4. Calculate \(\mu_k\).
5. Use the given relationship \(\mu_k = 0.7 \mu_s\) to solve for \(\mu_s\).
**Detailed Solution:**
1. **Calculate initial kinetic energy:**
\[ KE = \frac{1}{2} \cdot 40 \, \text{kg} \cdot (3 \, \text{m/s})^2 \]
\[ KE = \frac{1}{2} \cdot 40 \cdot
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