How to work out Joules needed to perform work

In summary: Ah, I think I understand it a bit better now. Thank you, this is very helpful.Note also that the human body is not an ideal physical machine. Physically, holding that wood block steady above your head involves no work, hence no expense of energy. But your muscles sure do burn calories while you do that!
  • #1
RetailPleb
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Hello all! First post here, and I apologize in advance if this question belongs elsewhere- I'll learn as I go.

I am interested in knowing more about energy and performing work. In particular, if I have an object I would like to lift, how much energy I need to expend to do that.

Take for example a block of wood which has a mass of 1 kilogram, and I'd like to pick it up off the ground, say, to a height of 2 meters. If I did my homework well enough, then my understanding is that this would require the same amount of energy whether I lifted it up over a period of 1 second or 1 minute, yes?

Work = Force * Distance. Work is expressed in joules, force in Newtons, and distance in meters.

So solving for work, and knowing the distance is 2 meters, that leaves force to work out.

A kilogram mass exerts a force of about 9.8 Newtons, since the Newton is equal to 1 kg⋅1 m/s2, and on Earth that would be about 9.8 m/s2.

So work = 9.8 * 2, or approximately 20 joules of energy, is this right? 20 joules is just under 5 calories. So the amount of energy I'd need to expend to lift this is, dare I say, nearly negligible, but still measurable.

If I could trouble the community to reassure me that I'm on the right track to working this out, or correct me if I've made a mistake somewhere in here, I'd be very grateful!
 
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  • #2
Yes, you have the work/force/distance relationship right and you're on the right track.
RetailPleb said:
20 joules is just under 5 calories. So the amount of energy I'd need to expend to lift this is, dare I say, nearly negligible, but still measurable.
Whether it's negligible or not somewhat depends on the context. If you're a typically sized and healthy human, yes, 20 Joules of mechanical work is negligible - you do more than that every time you get out of a chair. But if you were a mouse and your life depended on doing 20 Joules of work, "nearly negligible" is not what you would be thinking. Some physics experiments involve accurate measurements of energies that are many millions of times smaller.
 
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  • #3
Nugatory said:
Yes, you have the work/force/distance relationship right and you're on the right track.

Excellent! Thank you for the corroboration. And if you would humor me once more, if I got the above correct then this should also be right:

If I wanted to shove an adult human (let's say 70 kg) back about 5 meters, then...

Work = 686 N * 5m

3,430 Joules of energy is required, or about 800~ calories.

And this would be the same regardless of the time it took me, right? It's just a matter of a little bit of energy over a long period of time versus a lot of energy quickly?

Thanks again for indulging my silly questions!
 
  • #4
Bear in mind that if you are thinking of food calories, that what we call a calorie when we are talking about food is actually a kilocalorie. So 3430 Joules is only 0.8 food calories.
 
  • #5
phyzguy said:
Bear in mind that if you are thinking of food calories, that what we call a calorie when we are talking about food is actually a kilocalorie. So 3430 Joules is only 0.8 food calories.
Oooh, thank you! That is what I thought I was talking about. So actually this energy/effort amounts to a very puny morsel of a meal.
 
  • #6
RetailPleb said:
Oooh, thank you! That is what I thought I was talking about. So actually this energy/effort amounts to a very puny morsel of a meal.

Right. It takes a lot of work to burn off a 1000 calorie bacon cheeseburger.
 
  • #7
RetailPleb said:
Excellent! Thank you for the corroboration. And if you would humor me once more, if I got the above correct then this should also be right:

If I wanted to shove an adult human (let's say 70 kg) back about 5 meters, then...

Work = 686 N * 5m

3,430 Joules of energy is required, or about 800~ calories.
No, that's incorrect in general.

To lift a 70 kg human requires a force of 686 N. To raise that human by 5 meters takes 3430 Joules.

The force needed to shove the human a horizontally is not a fixed amount.
 
  • #8
SammyS said:
No, that's incorrect in general.

To lift a 70 kg human requires a force of 686 N. To raise that human by 5 meters takes 3430 Joules.

The force needed to shove the human a horizontally is not a fixed amount.

Ah, I think I understand it a bit better now. Thank you, this is very helpful.
 
  • #9
Note also that the human body is not an ideal physical machine. Physically, holding that wood block steady above your head involves no work, hence no expense of energy. But your muscles sure do burn calories while you do that!
 
  • #10
RetailPleb said:
If I wanted to shove an adult human (let's say 70 kg) back about 5 meters, then...
RetailPleb said:
Ah, I think I understand it a bit better now. Thank you, this is very helpful.
There's too many variables when trying to solve the "person moving backwards" problem (i.e. in what way did he move back, did he fall down, did he slide, etc) ... a better sample problem would be moving a solid box over a surface with a given coefficient of friction, and then finding the work needed to do complete that task.
 
  • #11
Comeback City said:
There's too many variables when trying to solve the "person moving backwards" problem (i.e. in what way did he move back, did he fall down, did he slide, etc) ... a better sample problem would be moving a solid box over a surface with a given coefficient of friction, and then finding the work needed to do complete that task.

Ah yes, I agree. I suppose a human was not a good example, but I also only wanted a general idea of what it might take. Precision was not my goal, only to make sure I had a good enough understanding of what is happening with such an event. But I appreciate your thoughtful response nonetheless!
 
  • #12
RetailPleb said:
Ah yes, I agree. I suppose a human was not a good example, but I also only wanted a general idea of what it might take. Precision was not my goal, only to make sure I had a good enough understanding of what is happening with such an event. But I appreciate your thoughtful response nonetheless!
No worries! Simple thought problems, and solving them, are great ways to continue learning.
 
  • #13
RetailPleb said:
Ah yes, I agree. I suppose a human was not a good example,
Absolutely. PF gets a regular stream of questions in which people try to analyse their sporting and fitness activities in terms of straightforward Physics. They nearly always leave dissatisfied with what PF has to say about it. People (in fact all plants and animals) are just too hard to look at that way.
 
  • #14
RetailPleb said:
So work = 9.8 * 2, or approximately 20 joules of energy, is this right?

Assuming you lifted it at a steady speed. If the object is accelerated, more force and hence more work is required.

20 joules is just under 5 calories. So the amount of energy I'd need to expend to lift this is, dare I say, nearly negligible, but still measurable.

20 joules of work is done on the object. but you do far more work than this because you raise the temperature of your body. It takes energy to do that, and you are the source.
 
  • #15
sophiecentaur said:
Absolutely. PF gets a regular stream of questions in which people try to analyse their sporting and fitness activities in terms of straightforward Physics. They nearly always leave dissatisfied with what PF has to say about it. People (in fact all plants and animals) are just too hard to look at that way.
Actually I have to confess, I'm gathering this information to work on a magic system for a TTRPG. I don't like the idea of "spell slots". The energy to use telekinesis and push someone back, for example, has to come from somewhere. I rather like the idea of spellcasters expending stores of body energy to do work, perhaps at a much higher cost than would otherwise be needed. I just want some small semblance of verisimilitude in how the magic works. Some sort of predictable, consistent laws. Of course working this way, the archetypal frail old wizard would probably be replaced with a more physically fit spellcaster, but hey, that's neat too I think.

I noticed there was a PF forum specifically for such sci-fi/fantasy topics, and when I get more of my thoughts put down in a coherent format I will venture over there, but for the time being I just need to make sure I understand the processes in how one does normal work before figuring out how a wizard would do it, haha.
 
  • #16
Mister T said:
Assuming you lifted it at a steady speed. If the object is accelerated, more force and hence more work is required.
20 joules of work is done on the object. but you do far more work than this because you raise the temperature of your body. It takes energy to do that, and you are the source.
Of course! Thank you for the clarification, much appreciated!
 
  • #17
DrClaude said:
Note also that the human body is not an ideal physical machine. Physically, holding that wood block steady above your head involves no work, hence no expense of energy. But your muscles sure do burn calories while you do that!
Indeed! And my body is certainly less efficient than I'd prefer, but I'm only aiming for ballpark estimates, not precise figures, so my back-of-the-envelope math should suffice for my needs.
 

Related to How to work out Joules needed to perform work

1. How do I calculate the amount of Joules needed to perform work?

To calculate the amount of Joules needed to perform work, you can use the formula W = F x d, where W represents work, F represents force, and d represents distance. You can also use the formula W = m x g x h, where m represents mass, g represents acceleration due to gravity, and h represents height.

2. What is the unit of measurement for Joules?

Joules (J) is the unit of measurement for energy and work. It is equivalent to one Newton-meter (N∙m) or one kilogram-meter squared per second squared (kg∙m²/s²).

3. Can I convert other units of energy to Joules?

Yes, you can convert other units of energy to Joules. For example, 1 calorie (cal) is equivalent to 4.184 Joules (J) and 1 British Thermal Unit (BTU) is equivalent to 1055.06 Joules (J).

4. How do I know how many Joules are needed for a specific task?

The amount of Joules needed for a specific task depends on the type of work being performed. For example, lifting an object to a certain height requires a specific amount of Joules, while running for a certain distance also requires a different amount of Joules. You can calculate the amount of Joules needed using the appropriate formula for the specific type of work.

5. What is the importance of knowing the amount of Joules needed for work?

Knowing the amount of Joules needed for work is important in order to understand and predict the amount of energy required for a specific task. This can help with planning and efficiency in various fields such as engineering, physics, and sports. It also allows for better understanding of energy conservation and the relationship between work and energy.

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