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I am holding a book of mass 0.5 kg in my right hand at a height of 1 m from ground level.

The book is lifted to a height of 1.5 m above ground and then brought down to its original height of 1 m. The book is at rest at 1.5 m and 1 m positions.

QUESTION

Is the person holding and moving the book losing energy, and why?

The answer I came up with is that the person should not loose any energy since the net work done on the hand by the book is 0. When person is lifting the book the book exerts a downward force on the hand while the hand is displaced upwards resulting in negative work. On the other hand, while bringing the book down the book exerts a downward force on the hand resulting in an equal positive work as displacement of hand is downwards. Thus net work done is 0 on the person and so person should not gain or lose energy.

Qmechanic
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Sunil
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2 Answers2

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Is the person holding and moving the book losing energy, and why?

Person holding and moving the book is actually losing energy all the time. The reason is that gravitational potential energy cannot be converted back to chemical energy stored in our muscles. Even when our body is seamlessly stationary, our muscles are constantly under motion on a very (very) small scale. This is the main reason why you cannot hold a book forever.

Although we are often trying to relate different concepts in physics with our intuition, which can help sometimes, you should not do this with work as defined in physics and effort done by our muscles. It will often lead to wrong conclusions.


The work energy theorem tells us that change in kinetic energy equals total work done on the object

$$\Delta K = W$$

The total work $W$ can be written as a sum of work done by gravitational force and work done by other forces (such as a person lifting a book)

$$\Delta K = W_g + W_\text{other} \tag 1$$

When you lift the book from $y_1$ to $y_2$, where positive $y$ direction points upwards and $y_1 < y_2$, the gravitational force is doing negative work on the book. Since book starts and ends with zero kinetic energy, the Eq. (1) for this case becomes

$$-mg(y_2-y_1) + W_\text{other} = 0 \qquad \rightarrow \qquad W_\text{other} = mg(y_2 - y_1) > 0$$

The work done by the person on the book $W_\text{other}$ is positive since $y_2 > y_1$, and the work done by the book on the person is negative which follows from the Newton's third law of motion. This is somehow intuitive.

When you put the book back down from $y_2$ to $y_1$, the kinetic energy difference is again zero and the work done by the person on the book is

$$mg(y_2-y_1) + W_\text{other} = 0 \qquad \rightarrow \qquad W_\text{other} = -mg(y_2-y_1) < 0$$

The work $W_\text{other}$ is now negative, which means the work done by the book on the person is positive. This is what physics tells us, and it is correct. What is not intuitive is the question "Where does this gained energy go?".

Major problem here is that we are trying to understand the work via effort done by our muscles. This is not equivalent and should be avoided. Although the book does positive work on the person when it goes down, the person still has to put some effort in preventing the book from free falling. This is done by constant contractions of our muscles which dissipate energy, i.e. they convert it to heat and maybe some other forms of energy.

When lifting the book, the effort done by our muscles converts to increase in gravitational potential energy of the book-Earth system. When putting the book down, the effort done by our muscles converts to decrease in gravitational potential energy of the book-Earth system. In both cases our muscles do effort, because gravitational potential energy is not converted back to chemical energy in our body (muscles).

Marko Gulin
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  • Yes in lifting it's positive on the book and therefore, book gains in energy. But what about work done on the person during lifting and descending? – Sunil Mar 07 '22 at 13:41
  • So, net work done on person's hand is 0. Thus, the hand does not gain or lose energy. Right? – Sunil Mar 07 '22 at 13:44
  • Also, chemical energy is converted to work while the book is being lifted. Right? But in downward descent, positive work is done on person's hand, so where does this positive work go? – Sunil Mar 07 '22 at 13:47
  • @Sunil The positive work goes to different processes in our body. But in the end, our muscles again use chemical energy to do some work - much less than when lifting the book, but they still have to do some work in absorbing the gravitational potential energy. The absorbed energy cannot be converted back to chemical energy, it is just not how our body works. – Marko Gulin Mar 07 '22 at 13:49
  • When you said "The positive work goes to different processes in our body. ", what would be these different processes in body? Please explain. Thankyou. – Sunil Mar 07 '22 at 13:51
  • @Sunil Muscles do effort by constant "up-down" movement. When you are putting the book down, your muscles internally constantly go up-down on a very small case. This is how the energy gained is dissipated in our muscles. – Marko Gulin Mar 07 '22 at 13:54
  • You meant energy gained by hand gets lost since the elastic PE in our muscles decreases? – Sunil Mar 07 '22 at 13:58
  • @Sunil I do not know how muscles are actually modelled, but yes, the energy is "dissipated" in muscles when lowering the book. It has to be! The energy cannot be destroyed, it can only change the form. Since the person does not gain kinetic or potential energy while lowering the book, the work done by the book on the person is positive and is dissipated by our muscles in a form of heat. Maybe to some other forms as well, but I do not know specifics. – Marko Gulin Mar 07 '22 at 14:04
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There is never loss or gain of energy , energy is always converted in different forms .

If I hold something a book which has x amount of joules as potential energy at the ground level . Now I take it in my hand raise it to a higher level . If u notice , that there is a force ( my potential energy ) being used to up to overcome the gravity and hold it at a higher level , with the gravity acting downwards. As a result of this ,a part of my potential energy gets converted to the the object's potential energy increasing it from "x" joules.

So when I raise it form 1m to 1.5m , I lose energy which gets converted to potential energy of the book .

Similarly when I bring it down to a 1m level , it should lose energy with a net 0 gain or loss of energy when looking at the entire situation.

Yet if u look at it more closely , u'll notice the person still does some work ( whether its negative or positive) , it does require some part of a person's energy to bring something down .

According to me the person should lose energy