Help with the work done problem

In summary, the conversation discusses a confusion regarding the work done on a load in a given problem. Some argue that the work done is W(x+y) while others argue it is just Wy. There is a request for clarification and different arguments are presented, including the definition of work and the relevance of relative directions of displacement and force.
  • #1
jeevankasin
3
0
Hi,

Some confusion with the attached problem. Some of us say that the work done on the load is W(x+y) while some others are saying that the WD on the load is just Wy since the motor moves the distance x and that work is done by the motor.

Could we have some clarifications?

Din
 

Attachments

  • Work Done Q.jpg
    Work Done Q.jpg
    28.7 KB · Views: 371
Physics news on Phys.org
  • #2
jeevankasin said:
Hi,

Some confusion with the attached problem. Some of us say that the work done on the load is W(x+y) while some others are saying that the WD on the load is just Wy since the motor moves the distance x and that work is done by the motor.

Could we have some clarifications?

Din

Welcome to the PF.

What are your arguments in favor of each choice?
 
  • #3
Option (A): One of us argues, since that is the distance the object moves, the WD should be that distance.

Option (B): The WD here is to bring the load up by a dist of y and multiplied by the force acting on it, which is the weight.

Force x Dist moved in the direction of the force

Option (D) : Since the object moves a horizontal distance of x followed by the vertical distance y, thus it W(x+y).
 
  • #4
What's the definition of work?

Do the relative directions of displacement and force matter?
 
  • #5
,

Thank you for reaching out and seeking clarification on this problem. The concept of work done can be confusing, so I am happy to provide some clarification.

In this scenario, both answers are technically correct depending on how you define the system and the work being done. Let me explain further:

First, let's define the system. In this problem, the system consists of the motor and the load. The motor is doing work on the load by moving it a distance x. However, the load is also moving a distance y due to external forces acting on it. So, in this case, the work done on the load can be calculated as W(x+y). This takes into account both the work done by the motor and the work done by external forces on the load.

On the other hand, if we only consider the work done by the motor, then the answer would be Wy. This is because the motor is only responsible for moving the load a distance x, and the external forces are responsible for the additional distance y. In this case, we are only looking at the work done by the motor, not the total work done on the load.

So, in summary, both answers are correct depending on the perspective and definition of the system. I hope this helps clarify the confusion. Please let me know if you have any further questions.
 

Related to Help with the work done problem

1. What is the work done problem?

The work done problem is a physics concept that involves calculating the amount of work (in joules) done on an object by a force. It is based on the formula W = F * d, where W is work, F is force, and d is displacement.

2. How do I solve a work done problem?

To solve a work done problem, you will need to know the values for force and displacement and plug them into the formula W = F * d. Make sure to use the correct units for each value (newtons for force and meters for displacement) and check your answer using the proper units for work (joules).

3. What is the difference between work done and power?

Work done and power are two related but distinct concepts in physics. Work done is a measure of the energy transferred to an object by a force over a certain distance, while power is the rate at which work is done or energy is transferred. In other words, power measures how quickly work is done, while work done measures the total amount of energy transferred.

4. Can you provide an example of a work done problem?

Sure! Let's say that a person pushes a box with a force of 50 newtons for a distance of 5 meters. To find the work done on the box, we would use the formula W = 50 N * 5 m = 250 joules. This means that 250 joules of energy was transferred to the box by the force of 50 newtons over a distance of 5 meters.

5. How is work done related to kinetic energy?

Work done is directly related to kinetic energy, which is the energy an object possesses due to its motion. The work-energy theorem states that the net work done on an object is equal to the change in its kinetic energy. In other words, work done can either increase or decrease an object's kinetic energy, depending on the direction of the force and the displacement.

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
346
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
17
Views
812
  • Introductory Physics Homework Help
Replies
9
Views
994
  • Introductory Physics Homework Help
Replies
5
Views
631
  • Introductory Physics Homework Help
3
Replies
94
Views
3K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
2K
Back
Top