Calculate the work as the MINIMUM amount of work

In summary, the weightlifter must apply a force of at least 980 N to lift the weight. The work done is the difference in potential energies at the two points.
  • #1
danago
Gold Member
1,123
4
Hey. Heres the question:

"A weightlifter lifts a 100kg mass 2.4m from the floor. He then holds the mass steady for 3 seconds before dropping it. What is the work done by the weightlifer in raising the mass?"

Well. I know mass is applying a force downward of 980N because of gravity, so for this question, since an acceleration of the mass is not stipulated, do i just calculate the work as the MINIMUM amount of work he would do? Which would be around 2352J, so for the answer, would i just say that he must do MORE THAN 2352J of work to lift the weight?
 
Physics news on Phys.org
  • #2
It doesn't matter what the acceleration of the mass is. If here were to provide a drastic acceleration for 1 meter, he would no longer need to push upwards because the mass would have been thrown up and going up at this moment.
 
  • #3
If an object is applying a weight force of 980N downwards, and i applied a force back on it of 980N upwards, the object would be stationary right?

So to make the mass move upwards, wouldn't it require that the weightlifer apply a force of MORE than the weight force of 980N? OR am i completely missing the point here
 
  • #4
No you do not need to apply a force more that 980N, just 980N will do.

Any force that is more than 980N will result in an aceleration of the weight, since there is net force. ( X - 980 = Net force, where X >980)

The minimum amount of work that is required, requires the weight to be traveling at constant velocity.(Net force = zero, force applied=weight)

Hope it helps. :)
 
  • #5
danago said:
If an object is applying a weight force of 980N downwards, and i applied a force back on it of 980N upwards, the object would be stationary right?

So to make the mass move upwards, wouldn't it require that the weightlifer apply a force of MORE than the weight force of 980N? OR am i completely missing the point here

Yes, you are completely missing the point! Work= force*distance. Any force, above 980 N, applied to accelerate the the mass so that it will move upward, will have to be "negated" by a force under 980 N to stop it. The work done by those will cancel. The only force that counts is the 980 N that must be applied upward to counteract gravity and keep the weight moving at constant speed.

Or, you could argue that the work done is the difference in potential energies at the two points- again, weight*distance.
 
  • #6
Ahhh ok i see now :P

Thanks
 

1. What is meant by "minimum amount of work" in this context?

In this context, "minimum amount of work" refers to the least amount of energy required to complete a task. It is often used in physics and engineering to describe the efficiency of a system.

2. How is the minimum amount of work calculated?

The minimum amount of work is calculated by finding the lowest possible energy required to perform a specific task. This can be done through various mathematical equations, such as the work-energy theorem or the principle of least action.

3. Can the minimum amount of work be negative?

No, the minimum amount of work cannot be negative. Work is defined as the product of force and displacement, and both of these values are always positive. Therefore, the minimum amount of work can only be zero or a positive value.

4. What factors affect the minimum amount of work?

The minimum amount of work is affected by several factors, including the force applied, the distance over which the force is applied, and the angle between the force and the displacement. Additionally, the properties of the materials involved and any external forces can also impact the minimum amount of work.

5. How is the concept of minimum amount of work relevant in real-world applications?

The concept of minimum amount of work is relevant in various real-world applications, such as designing efficient machines and processes, optimizing energy usage, and understanding the limits of human and animal capabilities. It is also used in fields like economics to analyze cost-efficiency and in sports to improve performance through efficient movements.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
4K
  • Introductory Physics Homework Help
Replies
5
Views
297
  • Introductory Physics Homework Help
Replies
26
Views
3K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
7
Views
965
  • Introductory Physics Homework Help
Replies
12
Views
3K
  • Introductory Physics Homework Help
Replies
8
Views
3K
  • Introductory Physics Homework Help
Replies
5
Views
8K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
10
Views
1K
Back
Top