How much total work is done in raising the ball?

  • Thread starter Thread starter Kiko:)
  • Start date Start date
  • Tags Tags
    Ball Work
Click For Summary
SUMMARY

The total work done in raising a 0.075 kg ball 1.33 m using a motorized vertical conveyor belt is calculated by considering both the work done by the conveyor and the work done against friction. The frictional force of 0.350 N opposes the motion, and the work done against this friction is significant. The formula W = Fd is applied, where the total work done equals the work done by the conveyor minus the work done by friction, leading to the conclusion that the total work done is 0.099 J - 0.465 J, resulting in a total work of -0.366 J.

PREREQUISITES
  • Understanding of Newton's laws of motion
  • Familiarity with the work-energy principle
  • Knowledge of basic physics formulas, specifically W = Fd
  • Concept of frictional forces and their impact on motion
NEXT STEPS
  • Study the work-energy theorem in classical mechanics
  • Learn about the effects of friction on moving objects
  • Explore the concept of net work and its implications in physics
  • Investigate the role of motorized systems in kinetic sculptures
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in understanding the principles of work and energy in mechanical systems, particularly in the context of kinetic sculptures and motorized devices.

Kiko:)
Messages
16
Reaction score
0
A 0.075Kg ball in a kinetic sculpture is raised 1.33m above the ground by a motorized vertical conveyor belt. A constant frctional force of 0.350N acts in the direction opposite the conveyor belt. How much total work is done in raising the ball?

This confuses me quite a bit. I understand some of it. Could someone help me please? :)
 
Physics news on Phys.org
How would you normally find work done? What is the total energy of the sculpture?
 
rock.freak667 said:
How would you normally find work done? What is the total energy of the sculpture?

Use the formula W=Fd. And that's the problem I'm having is trying to find the total energy of the sculpture.
 
Kiko:) said:
Use the formula W=Fd. And that's the problem I'm having is trying to find the total energy of the sculpture.

the question says

A 0.075Kg ball in a kinetic sculpture is raised 1.33m above the ground

how much energy does it have at 1.33m above the ground?
 
.099?
 
I'm sorry I'm not trying to sound stupid. I'm just trying to understand this question!
 
Kiko:) said:
.099?

I don't have a calculator but it is around that. So this is the total amount of energy needed to lift the object 1.33m above the ground.

To lift the object, the total work done is the sum of the work done in lifting the object and what else?
 
rock.freak667 said:
To lift the object, the total work done is the sum of the work done in lifting the object and what else?

So then .345N force is acting on it and therefore since there is a force acting the opposite way there is no work that is being done on the ball in the kinetice sculpture. But there is work acting on the conveyor belt?
 
Kiko:) said:
So then .345N force is acting on it and therefore since there is a force acting the opposite way there is no work that is being done on the ball in the kinetice sculpture. But there is work acting on the conveyor belt?

right but the 0.345N does frictional work.

So the total work done = work by conveyor + work done by friction.

What is the work done by friction equal to? (see your 2nd post)
 
  • #10
rock.freak667 said:
right but the 0.345N does frictional work.

So the total work done = work by conveyor + work done by friction.

What is the work done by friction equal to? (see your 2nd post)


Use the equation W=Fd or .099?

So what your saying is to find the work by the conveyor and then add that to work done by friction (.350N) which then will equal the total work done in raising the ball?
 
  • #11
Kiko:) said:
Use the equation W=Fd or .099?

So what your saying is to find the work by the conveyor and then add that to work done by friction (.350N) which then will equal the total work done in raising the ball?

You have the right idea, just jumbled.

Total work done= work done by conveyor+ work done by friction.

You know the left side is 0.099. The work done by friction is Fd. So what is 'd' and the frictional work?
 
  • #12
rock.freak667 said:
You have the right idea, just jumbled.

Total work done= work done by conveyor+ work done by friction.

You know the left side is 0.099. The work done by friction is Fd. So what is 'd' and the frictional work?

Ah, so 0.099 = ______ + .350N x 1.33m?

0.099 + .465 = .564J?
 
  • #13
Kiko:) said:
Ah, so 0.099 = ______ + .350N x 1.33m?

0.099 + .465 = .564J?

yes, just change '+' to '-', I forgot to compensate with signs.
 
  • #14
rock.freak667 said:
yes, just change '+' to '-', I forgot to compensate with signs.

so .099 - .465 = 366J? And tha would be how much total work is done in raising the ball.
 
  • #15
Right?
 
  • #16
Kiko:) said:
so .099 - .465 = 366J? And tha would be how much total work is done in raising the ball.

No the previous answer was correct. It was supposed to be this

0.099 = ______ - .350N x 1.33m
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
2K
Replies
43
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 29 ·
Replies
29
Views
11K
  • · Replies 19 ·
Replies
19
Views
4K