### 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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Description: ### 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\

Given:mass of crate, m = 40 kgco-efficient of kinetic friction be μk co-efficient of static friction…

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(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?

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Newton's First Law of Motion

If a very smooth ball is rolled down an inclined surface with zero friction, the ball will continue to roll indefinitely for an infinite time, in an attempt to reach to the same height on the opposite hill!

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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

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