Why is small work done always taken as $dW=F\cdot dx$ and not $dW=x\cdot dF$ ?

Mariah Sparks
2022-07-23
Answered

Why is small work done always taken as $dW=F\cdot dx$ and not $dW=x\cdot dF$ ?

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Steven Bates

Answered 2022-07-24
Author has **15** answers

You can see this from the second Newton's law:

$m\ddot{\mathbf{x}}=\mathbf{F}(\mathbf{x})$

Now I would like to integrate this equation of motion with respect to time, to arrive at the energy conservation. To do so I multiply both sides with $\dot{\mathbf{x}}$:

$m\ddot{\mathbf{x}}\cdot \dot{\mathbf{x}}=\mathbf{F}(\mathbf{x})\cdot \dot{\mathbf{x}}$

and finally integrate:

$m\int dt\ddot{\mathbf{x}}\cdot \dot{\mathbf{x}}=\int dt\mathbf{F}(\mathbf{x})\cdot \dot{\mathbf{x}}$

The l.h.s. gives me the kinetic energy. The r.h.s. gives me exactly the integral in question:

$\frac{1}{2}m{\dot{\mathbf{x}}}^{2}=\int d\mathbf{x}\cdot \mathbf{F}(\mathbf{x})$So the work done by the force is the kinetic energy of the particle (up to an integration constant representing its total energy).

$m\ddot{\mathbf{x}}=\mathbf{F}(\mathbf{x})$

Now I would like to integrate this equation of motion with respect to time, to arrive at the energy conservation. To do so I multiply both sides with $\dot{\mathbf{x}}$:

$m\ddot{\mathbf{x}}\cdot \dot{\mathbf{x}}=\mathbf{F}(\mathbf{x})\cdot \dot{\mathbf{x}}$

and finally integrate:

$m\int dt\ddot{\mathbf{x}}\cdot \dot{\mathbf{x}}=\int dt\mathbf{F}(\mathbf{x})\cdot \dot{\mathbf{x}}$

The l.h.s. gives me the kinetic energy. The r.h.s. gives me exactly the integral in question:

$\frac{1}{2}m{\dot{\mathbf{x}}}^{2}=\int d\mathbf{x}\cdot \mathbf{F}(\mathbf{x})$So the work done by the force is the kinetic energy of the particle (up to an integration constant representing its total energy).

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What is the difference b/n spontaneity criterial of simple and non simple systems in terms of gibbs and helmholtz free energies?

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The first law of thermodynamics

$\text{d}Q=\text{d}U+p\text{d}V-\mu \text{d}N$

is equivalent to saying: Q is a differentiable real valued function of U,V and N, so that we can write $\text{d}Q=\frac{\mathrm{\partial}Q}{\mathrm{\partial}U}\text{d}U+\frac{\mathrm{\partial}Q}{\mathrm{\partial}V}\text{d}V+\frac{\mathrm{\partial}Q}{\mathrm{\partial}N}\text{d}N$, and $\frac{\mathrm{\partial}Q}{\mathrm{\partial}U}=1$. Also, we define $p=\frac{\mathrm{\partial}Q}{\mathrm{\partial}V}$ and $-\mu =\frac{\mathrm{\partial}Q}{\mathrm{\partial}N}$

Is that it?

$\text{d}Q=\text{d}U+p\text{d}V-\mu \text{d}N$

is equivalent to saying: Q is a differentiable real valued function of U,V and N, so that we can write $\text{d}Q=\frac{\mathrm{\partial}Q}{\mathrm{\partial}U}\text{d}U+\frac{\mathrm{\partial}Q}{\mathrm{\partial}V}\text{d}V+\frac{\mathrm{\partial}Q}{\mathrm{\partial}N}\text{d}N$, and $\frac{\mathrm{\partial}Q}{\mathrm{\partial}U}=1$. Also, we define $p=\frac{\mathrm{\partial}Q}{\mathrm{\partial}V}$ and $-\mu =\frac{\mathrm{\partial}Q}{\mathrm{\partial}N}$

Is that it?

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If you want to cook in water at ${150}^{\circ}C$, you need a pressure cooker that can with stand necessary pressure. a) What pressure is required for ht boiling point of water to be this high? b) If the lid of the pressure cooker is a disk 25.0 cm in diameter, what force must it be able to withstand, at this pressure?

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What calorimetry approach is this?

Question:

Find the final tempertature of the system when $5gm$ of steam at ${100}^{\circ}C$ is mized with $31.5gm$ ice at $-{20}^{\circ}C$

Constants:

Specific heat of ice and specific heat of steam is $0.5cal/gm{/}^{\circ}C$

Specific heat of water is $1cal/gm{/}^{\circ}C$

Latent heat of fusion is $80cal/gm$

Latent heat of vaporization is $540cal/gm$

My attempt:

I usually approach these problems by setting the final temperature as $T$. Then, I equate heat lost by steam to heat gained by ice. I have been taught this method and I am adept at it.

My problem: The solution given in my workbook is:

Heat released when $1g$ of steam at ${100}^{\circ}C$ is cooled to $-{20}^{\circ}C$ is $730cal$. Now, this heat is given to be given to $1+6.3g=7.3g$ of ice. This raises the temperature by ${10}^{\circ}C$

This is a new method for me, and I am unable to understand it. I wish to know:

How does this method work? In what calorimetry situations can this method be applied?

NOTE: The final answer is ${10}^{\circ}C$

Question:

Find the final tempertature of the system when $5gm$ of steam at ${100}^{\circ}C$ is mized with $31.5gm$ ice at $-{20}^{\circ}C$

Constants:

Specific heat of ice and specific heat of steam is $0.5cal/gm{/}^{\circ}C$

Specific heat of water is $1cal/gm{/}^{\circ}C$

Latent heat of fusion is $80cal/gm$

Latent heat of vaporization is $540cal/gm$

My attempt:

I usually approach these problems by setting the final temperature as $T$. Then, I equate heat lost by steam to heat gained by ice. I have been taught this method and I am adept at it.

My problem: The solution given in my workbook is:

Heat released when $1g$ of steam at ${100}^{\circ}C$ is cooled to $-{20}^{\circ}C$ is $730cal$. Now, this heat is given to be given to $1+6.3g=7.3g$ of ice. This raises the temperature by ${10}^{\circ}C$

This is a new method for me, and I am unable to understand it. I wish to know:

How does this method work? In what calorimetry situations can this method be applied?

NOTE: The final answer is ${10}^{\circ}C$

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