**Title: Work Done on a Bag on an Inclined Conveyor Belt****Scenario:**A piece of luggage is being loaded onto an airplane using an inclined conveyor belt. The bag, with a mass of 10.0 kg, travels a distance \( d = 6.50 \, \text{m} \) up the conveyor belt at a constant speed without slipping.**Task:**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 (\( W_{\text{net}} \))**Objective:**Match the work done by each force to the appropriate energy value.**Diagram Explanation:**The diagram illustrates a side view of an inclined plane with angle \( \theta \), showing the luggage moving a distance \( d \) along the plane.**Energy Values (to Match):**- 638 J- 552 J- \(-638 \, \text{J}\)- \(-552 \, \text{J}\)- 0 J- 319 J- \(-319 \, \text{J}\)**Match Energy Values:**- \( W_g \)- \( W_N \)- \( W_f \)- \( W_{\text{conveyor}} \)- \( W_{\text{net}} \)**Note:**This exercise aims to enhance understanding of work and energy concepts applied to an object moving on an inclined plane.

Name: **Title: Work Done on a Bag on an Inclined Conveyor Belt****Scenario:
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Description: **Title: Work Done on a Bag on an Inclined Conveyor Belt****Scenario:**A piece of luggage is being loaded onto an airplane using an inclined conveyor belt. The bag, with a mass of 10.0 kg, travels a distance \( d = 6.50 \, \text{m} \) up the conveyo

The work done by the force of gravity is Wg = mgh = mgl sin60° =(10 kg ) × (- 9.8 m/s 2 ) × (6.50 m) √3/2 = - 551.65 =( - 552 J ) The work done by the force of gravity is. - 552 JThe work done by the normal force WN = FN sN = mg cos60° (0) = 0 The work done by the normal force is 0 c) The work of the friction force Wf = mg sin 60 × L =10 kg × 9.8 N/kg × √3/2 × 6.50 = 552 J The workdone by the conyer belt is as the body is going at constant speed the force of the tape is equal to the force of friction Wc = W f Wf = mg sin 60 × L =10 kg × 9.8 N/kg × √3/2 × 6.50 = 552 J the workdone by the net force Wnet = 0 Since, there is no change in the kinetic energy of the block the workdone by the net force Is 0

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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 10.0 kg, travels d = 6.50 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 (Wf) • the conveyor belt (Wconveyor ) the net force (Wnet) Match the work done by each force to the appropriate energy value. 638 J 552 J -638 J -552 J OJ 319 J -319 J Wg WN

A piece of luggage is being loaded onto an airplane by way of an inclined conveyor belt. The bag, which has a mass of 10.0 kg, travels d = 6.50 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 (Wf) • the conveyor belt (Wconveyor ) the net force (Wnet) Match the work done by each force to the appropriate energy value. 638 J 552 J -638 J -552 J OJ 319 J -319 J Wg WN

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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Topic Video

Question

**Title: Work Done on a Bag on an Inclined Conveyor Belt**

**Scenario:**
A piece of luggage is being loaded onto an airplane using an inclined conveyor belt. The bag, with a mass of 10.0 kg, travels a distance \( d = 6.50 \, \text{m} \) up the conveyor belt at a constant speed without slipping.

**Task:**
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 (\( W_{\text{net}} \))

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

**Diagram Explanation:**
The diagram illustrates a side view of an inclined plane with angle \( \theta \), showing the luggage moving a distance \( d \) along the plane.

**Energy Values (to Match):**
- 638 J
- 552 J
- \(-638 \, \text{J}\)
- \(-552 \, \text{J}\)
- 0 J
- 319 J
- \(-319 \, \text{J}\)

**Match Energy Values:**

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

**Note:**
This exercise aims to enhance understanding of work and energy concepts applied to an object moving on an inclined plane.

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