Question on work power and energy [duplicate]

Question

A person holds a bundle of grass over his head for 30 minutes and gets tired. Has he done some work or not please explain me?

Answer 1

I'll go ahead and go the shocking route of saying, yes, you have done work.

But not on the bundle of grass (?), and not on the macroscopic level, only on the microscopic level.

We'll need some basic musculoskeletal biophysics to understand this (see diagram below). Your muscle cells contain interlaced strands of actin and myosin proteins. When your muscle is contracted, these proteins are constantly grabbing and pulling on each other, then letting go, then grabbing and pulling again. Each time they do this, they receive an input of chemical energy from an ATP molecule that allows for their repeated conformational changes. Each time they return to their original position, the energy they gained is transferred through a million molecular collisions with their surrounding cytoplasm into heat, analogous to a block repeatedly pushed up a hill and allowed to slide back down to a stop at the bottom, the potential energy it had transferred to heat. Each of those molecular collisions therefore represents a tiny bit of microscopic work (the actin/myosin complex exerts a tiny $W = \vec{F}\cdot \vec{d}$ on the molecules it collides with as it returns to its un-energized state). This work is not done on the bundle of grass, mind you, but on the various molecules constituting your muscle cells.

So, to sum up, when you hold an object over your head, you are using chemical energy to do work in a distributed microscopic way on the molecules of your muscle cells, but not in a macroscopic way, and not on the object itself, which does not gain any energy (since $W = \vec{F}\cdot \vec{d}$, and $d = 0$ for the object you're holding). This work that you do on your own molecules is responsible for the decrease in chemical energy that you experience, and, in turn, the increase in thermal energy (heat) that you can feel whenever you contract your muscles.

Answer 2

In physics, the work is done if the displacement takes place in the direction of the force applied. Mathematically, in vector form
$$W=\vec F\cdot \vec d$$ Where, $\vec F$ is force vector & $\vec d$ is displacement vector

The weight is acting vertically downward & there is no displacement in the same direction hence work done $W=\vec F\cdot (\vec 0)=0$

Answer 3

Work is defined in physics as transfer of energy into useful(into your system of study) mechanical motion(kinetic energy) of an object. Notice the term useful, means it goes into something you can immediately see, like a car or a bus or a block in motion. So, in this case when you are standing with a block, you're essentially doing no work as defined by physics whose mathematical expression is $\vec{F}\cdot\vec{d}$, on the block as the block's state of motion isn't changed. However, you are losing energy, by radiating heat into EM waves, into the atmosphere by collision of air molecules with you. This is also work, as this is transfer of energy too, but this is kind of useless, as you don't immediately see it and the air and the universe isn't your system of study.

When you'll study thermodynamics, you'll see work as transfer of energy into air molecules' kinetic energy and that'll be counted as useful, since then those chambers will be your system of study. So, to define work, we first define a system and you've to change its kinetic energy.

You can even call transfer of energy into the EM field 'work', when the electromagnetic field is your system of study, but we generally don't do that and simply call it energy transfer, since it isn't kinetic energy.

For more info about relation of these kind of issues, I recommend you to read Physics Vol. 1 by Resnick, Halliday and Krane, where they discuss(in sometimes paper linked therein), why students at times fail to understand that the relation between the first law of thermodynamics and what they studied in newtonian mechanics.

So, in summary, work is transfer of energy to kinetic energy of your system of study, which can be air molecules or a block, from whatever form it was present in before and to answer your question, no the person hasn't done any work, because the block which is your system of study, didn't change its kinetic energy.

EDIT

Physics : How to work?

1. Choose your system of study.
2. Do whatever you want but if you feel tired go to step 3.

3. Did you change the Kinetic energy of the system?

   • If yes, you've done work on the system.
   • If no, you've done no work on the system.

*(You might still have done work elsewhere, just not on your system.)

** (To see where you've done work, isn't so simple. You've to choose new system wisely.)

Answer 4

It is true that work is not done when you hold an object without any motion. But that is true in ideal conditions. When you hold a heavy object even for a few minutes you get tired. Not only that even if you stand idle you get tired. Then is it that laws of physics are violated??

No. Even if you stand idle, your body does work to stabilize you, that is you do not fall down. Now if you catch a heavy weight your muscles do greater amount of work to stabilize you and so you spend greater amount of energy. This is the reason a person gets tired when he holds some heavy object.