A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, travels d = 4.50 m up the conveyor belt at a constant speed without slipping. If the conveyor belt is inclined at an angle 0 = 60.0°, calculate the work done on the bag by • the force of gravity (Wg) • the normal force (WN) • the friction force (W;) • the conveyor belt (Wconveyor ) • the net force (Wnet) Match the work done by each force to the appropriate energy value. 662 J 573 J -662 J -573 J OJ 331 J -331 J Answer Bank WN Wconveyor

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A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, travels \( d = 4.50 \, \text{m} \) up the conveyor belt at a constant speed without slipping.

If the conveyor belt is inclined at an angle \( \theta = 60.0^\circ \), calculate the work done on the bag by:

- the force of gravity (\( W_g \))
- the normal force (\( W_N \))
- the friction force (\( W_f \))
- the conveyor belt (\( W_{\text{conveyor}} \))
- the net force (\( W_{\text{net}} \))

**Match the work done by each force to the appropriate energy value:**

- \( 662 \, \text{J} \)
- \( 573 \, \text{J} \)
- \( -662 \, \text{J} \)
- \( -573 \, \text{J} \)
- \( 0 \, \text{J} \)
- \( 331 \, \text{J} \)
- \( -331 \, \text{J} \)

**Answer Bank:**

- \( W_N \)
- \( W_{\text{conveyor}} \)
- \( W_{\text{net}} \)
- \( W_g \)
- \( W_f \)

The diagram shows a conveyor belt inclined at angle \( \theta \) with the horizontal. Two rollers support the belt where a piece of luggage moves a distance \( d \) along the inclined plane.
Transcribed Image Text:A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 15.0 kg, travels \( d = 4.50 \, \text{m} \) up the conveyor belt at a constant speed without slipping. If the conveyor belt is inclined at an angle \( \theta = 60.0^\circ \), calculate the work done on the bag by: - the force of gravity (\( W_g \)) - the normal force (\( W_N \)) - the friction force (\( W_f \)) - the conveyor belt (\( W_{\text{conveyor}} \)) - the net force (\( W_{\text{net}} \)) **Match the work done by each force to the appropriate energy value:** - \( 662 \, \text{J} \) - \( 573 \, \text{J} \) - \( -662 \, \text{J} \) - \( -573 \, \text{J} \) - \( 0 \, \text{J} \) - \( 331 \, \text{J} \) - \( -331 \, \text{J} \) **Answer Bank:** - \( W_N \) - \( W_{\text{conveyor}} \) - \( W_{\text{net}} \) - \( W_g \) - \( W_f \) The diagram shows a conveyor belt inclined at angle \( \theta \) with the horizontal. Two rollers support the belt where a piece of luggage moves a distance \( d \) along the inclined plane.
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