An arrangement of two pulleys, as shown in the figure, is used to lift a 77.5-kg crate a distance of 2.78 m above the starting point. Assume the pulleys and rope are ideal and that all rope sections are essentially vertical. P kg where P = 77.5. What is the change in the potential energy of the crate when it is lifted a distance of 2.78 m? kJ

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An arrangement of two pulleys, as shown in the figure, is used to lift a 77.5-kg crate a distance of 2.78 m above the starting point. Assume the pulleys and rope are ideal and that all rope sections are essentially vertical.

*Diagram description:* The image shows a pulley system with two pulleys. A rope runs over the pulleys, lifting a crate labeled with "P kg," where P = 77.5. An arrow labeled "F" indicates the force applied to lift the crate.

What is the change in the potential energy of the crate when it is lifted a distance of 2.78 m?

_____ kJ
Transcribed Image Text:An arrangement of two pulleys, as shown in the figure, is used to lift a 77.5-kg crate a distance of 2.78 m above the starting point. Assume the pulleys and rope are ideal and that all rope sections are essentially vertical. *Diagram description:* The image shows a pulley system with two pulleys. A rope runs over the pulleys, lifting a crate labeled with "P kg," where P = 77.5. An arrow labeled "F" indicates the force applied to lift the crate. What is the change in the potential energy of the crate when it is lifted a distance of 2.78 m? _____ kJ
### Pulley System and Work Calculation

#### Overview
This exercise involves a pulley system intended to lift a crate. The system consists of two pulleys, with the setup designed to lift a 77.5 kg crate vertically by a distance of 2.78 meters. Assume that both the pulleys and rope are ideal, meaning there is no friction and all sections of the rope are vertical.

#### Diagram Explanation
- **Components:**
  - The setup includes two pulleys.
  - A crate with a mass denoted as \( P \) kilograms, where \( P = 77.5 \).
  - A force \( F \) is applied to lift the crate.

The diagram shows the two pulleys with a rope running over them. One end of the rope is attached to the crate, while the other end is subjected to an external force \( F \) directed downwards. 

#### Task
Calculate the amount of work required to lift the crate a vertical distance of 2.78 meters.

#### Work Calculation
- **Formula:** Work done is calculated using the formula:

  \[
  \text{Work} = \text{Force} \times \text{Distance}
  \]

- **Gravitational Force:** The force required to lift the crate can be determined by:

  \[
  F = m \times g
  \]

  where:
  - \( m = 77.5 \) kg (mass of the crate)
  - \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity)

- **Distance:** 2.78 meters

- **Work Done:**

  \[
  \text{Work} = F \times \text{Distance}
  \]

  \[
  \text{Work} = (77.5 \times 9.81) \times 2.78
  \]

- **Final Result:** Calculate the above expression to find the work in kilojoules (convert joules to kilojoules by dividing by 1000).

This exercise helps understand the fundamentals of mechanical advantage and work in a pulley system.
Transcribed Image Text:### Pulley System and Work Calculation #### Overview This exercise involves a pulley system intended to lift a crate. The system consists of two pulleys, with the setup designed to lift a 77.5 kg crate vertically by a distance of 2.78 meters. Assume that both the pulleys and rope are ideal, meaning there is no friction and all sections of the rope are vertical. #### Diagram Explanation - **Components:** - The setup includes two pulleys. - A crate with a mass denoted as \( P \) kilograms, where \( P = 77.5 \). - A force \( F \) is applied to lift the crate. The diagram shows the two pulleys with a rope running over them. One end of the rope is attached to the crate, while the other end is subjected to an external force \( F \) directed downwards. #### Task Calculate the amount of work required to lift the crate a vertical distance of 2.78 meters. #### Work Calculation - **Formula:** Work done is calculated using the formula: \[ \text{Work} = \text{Force} \times \text{Distance} \] - **Gravitational Force:** The force required to lift the crate can be determined by: \[ F = m \times g \] where: - \( m = 77.5 \) kg (mass of the crate) - \( g = 9.81 \, \text{m/s}^2 \) (acceleration due to gravity) - **Distance:** 2.78 meters - **Work Done:** \[ \text{Work} = F \times \text{Distance} \] \[ \text{Work} = (77.5 \times 9.81) \times 2.78 \] - **Final Result:** Calculate the above expression to find the work in kilojoules (convert joules to kilojoules by dividing by 1000). This exercise helps understand the fundamentals of mechanical advantage and work in a pulley system.
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