Friction of conveyor belt on suitcase

Click For Summary
SUMMARY

The discussion focuses on calculating the distance a 10 kg suitcase is dragged on a conveyor belt moving at 2 m/s, considering static and kinetic friction coefficients of 0.5 and 0.3, respectively. Participants emphasize using the work-energy principle, specifically the equation E_{kinetic} = F \cdot \Delta s, to relate kinetic energy to the work done by friction. The conversation clarifies that the energy for moving the suitcase comes from the motor driving the conveyor belt, not from the suitcase's kinetic energy. Participants conclude that the friction force does work on the suitcase, converting energy into kinetic energy until it matches the belt's speed.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with work-energy principles
  • Knowledge of static and kinetic friction coefficients
  • Ability to perform free body diagrams (FBDs)
NEXT STEPS
  • Study the work-energy theorem in detail
  • Learn how to calculate distances using friction forces
  • Explore free body diagram techniques for dynamic systems
  • Investigate the relationship between kinetic energy and work done by friction
USEFUL FOR

Students in physics or engineering, particularly those studying dynamics and energy transfer, as well as educators looking for practical examples of work-energy applications.

habibclan
Messages
55
Reaction score
0

Homework Statement



A baggage handler drops your 10 kg suitcase onto a conveyer belt running at 2 m/s. The materials are such that us= 0.5 and uk=0.3 [coefficients of static and kinetic friction]. How far is your suitcase dragged before it is riding smoothly on the belt?


Homework Equations



fs= usmg
fk= ukmg

The Attempt at a Solution



I don't know how to set up the problem because I don't know how to define the force applied by the baggage handler. Any help would be greatly appreciated! Thanks in advance! =)
 
Physics news on Phys.org
Draw a picture and identify the friction force then use an energy balance to find the traveled distance.

\frac{1}{2}mv^2 = \mu_{total}mg \Delta x
 
Last edited:
I really have no idea. I was able to do all the other questions in this chapter, just not this one. I would say that the friction is in the direction of the motion of the conveyer belt because due to friction, the luggage is able to stick to the conveyer belt. Can you please guide me as to how to set up the problem!
 
dirk_mec1 said:
Draw a picture and identify the friction force then use an energy balance to find the traveled distance.

\frac{1}{2}mv^2 = \mu_{total}mg \Delta x

How did you get the friction equation with delta x in it? I've never seen that before.
 
habibclan said:
How did you get the friction equation with delta x in it? I've never seen that before.

Yes you have it's just:

E_{kinetic} = F \cdot \Delta s

with s the distance
 
dirk_mec1 said:
Yes you have it's just:

E_{kinetic} = F \cdot \Delta s

with s the distance

Oh okay! This is the work equation. Thanks a lot =). But logically, what happens? When the guy puts the suitcase onto the belt, from this equation, it means that the kinetic energy of the belt is converted into energy for friction?
 
habibclan said:
Oh okay! This is the work equation. Thanks a lot =). But logically, what happens? When the guy puts the suitcase onto the belt, from this equation, it means that the kinetic energy of the belt is converted into energy for friction?
No the guys drops the suitcase that's something else then putting it on the belt.

Imagine yourself a suitcase dropping on a conveyer belt the suitcase 'slides' a distance and then just moves with the belt, get the picture?
 
dirk_mec1 said:
No the guys drops the suitcase that's something else then putting it on the belt.

Imagine yourself a suitcase dropping on a conveyer belt the suitcase 'slides' a distance and then just moves with the belt, get the picture?

That I get. But I feel so stupid coz I can't understand how Efric=Kinetic energy. Is it that the kinetic energy of the belt converts itself into friction energy to hang onto the luggage? Sorry for bugging you so much :$.
 
habibclan said:
That I get. But I feel so stupid coz I can't understand how Efric=Kinetic energy. Is it that the kinetic energy of the belt converts itself into friction energy to hang onto the luggage?

Hi habibclan! :smile:

Think of the belt as an infinite source of energy …

… rather like the Earth is an "infinite" source of mass … when you bounce a ball off a wall, you tend to assume that the wall is fixed to the Earth and the Earth doesn't move! :smile:

The belt will go on moving at the same speed no matter how kmany cases are put on it … the speed is determined by (a) the engine running it, and (b) the person who wrote the question.

Forget the KE of the belt.

When you do FBDs, you do them for only one body at a time.

(You could do an FBD for the belt, but you'd have to pult the belt motor into it, and you've no idea what that's doing! :rolleyes: )

In this case do an FBD, or conservation equation, or work-energy equation, for the luggage only.

You know it starts with zero horizontal velocity, and there's only one horizontal force on it, which is constant until it reaches the velocity of the belt.

You can either do a Newton's second law equation … which has the disadvantage of giving you a differential equation to solve :redface: , or, preferably in this case, use the Work-energy equation that dirk_mec1 suggested, which tells you that the KE of the luggage increases because of the work done on it by the friction force. :smile:
 
  • #10
tiny-tim said:
Hi habibclan! :smile:

Think of the belt as an infinite source of energy …

… rather like the Earth is an "infinite" source of mass … when you bounce a ball off a wall, you tend to assume that the wall is fixed to the Earth and the Earth doesn't move! :smile:

The belt will go on moving at the same speed no matter how kmany cases are put on it … the speed is determined by (a) the engine running it, and (b) the person who wrote the question.

Forget the KE of the belt.

When you do FBDs, you do them for only one body at a time.

(You could do an FBD for the belt, but you'd have to pult the belt motor into it, and you've no idea what that's doing! :rolleyes: )

In this case do an FBD, or conservation equation, or work-energy equation, for the luggage only.

You know it starts with zero horizontal velocity, and there's only one horizontal force on it, which is constant until it reaches the velocity of the belt.

You can either do a Newton's second law equation … which has the disadvantage of giving you a differential equation to solve :redface: , or, preferably in this case, use the Work-energy equation that dirk_mec1 suggested, which tells you that the KE of the luggage increases because of the work done on it by the friction force. :smile:

I get it now! THanks so much! So basically this is one of the rare cases where thermal energy is converted into kinetic energy. I was reading my textbook today and it said that usually its KE--> thermal, but I guess this is an exception to that case!
 
  • #11
habibclan said:
So basically this is one of the rare cases where thermal energy is converted into kinetic energy. I was reading my textbook today and it said that usually its KE--> thermal, but I guess this is an exception to that case!

No no no … the energy comes from the motor which drives the belt.

Some of that energy moves the luggage, and some is converted into thermal energy.

Friction is not thermal … the useful work done by friction is converted into KE … the wasted work becomes thermal energy. :smile:
 
  • #12
tiny-tim said:
No no no … the energy comes from the motor which drives the belt.

Some of that energy moves the luggage, and some is converted into thermal energy.

Friction is not thermal … the useful work done by friction is converted into KE … the wasted work becomes thermal energy. :smile:

Thanks a lot!
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 6 ·
Replies
6
Views
9K
  • · Replies 27 ·
Replies
27
Views
4K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 11 ·
Replies
11
Views
16K
Replies
7
Views
4K
Replies
2
Views
5K
  • · Replies 3 ·
Replies
3
Views
11K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 24 ·
Replies
24
Views
5K