# Tag Info

## New answers tagged work

0

Your muscles / musculo-skeletal system is producing ATP for metabolic purposes, and your body releases much of this byproduct energy in the form of radiation if your body is cooler than the surrounding ambient temperature, and otherwise through evaporation. Like others have mention, some is "lost" in the form of sound and friction.

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Make it simple. If a mass of your weight fell down the height of three flights of stairs through the air, when it landed where would the kinetic energy accumulated by falling go? moving the earth for conservation of momentum dissipated in heat on the ground deformation of the matter of the weight sound That is why humans invented the stairs, to ...

8

Yes, once you are standing still at the bottom of the stairs then all of the potential energy you had at the top of the stairs has been transformed into heat (due to friction and heat produced by your body) and sound waves. And even the energy in the sound waves ends up as heat once it has been absorbed by the walls, floor, ceiling etc.

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The heat is predominently generated in your muscles. A more direct conversion of potential energy to heat is when a person is sliding down a pole to get to a lower floor quickly. With sufficient friction the descent is at a constant velocity instead of accelerating. In muscle there are structures that slide along each other. Muscle contraction is those ...

5

The potential energy of your body starting at a height is gradually lost with each step. Your body is transferring this potential energy into the stretching and flexing of your muscles, and the heat created by this. Some of the energy is also lost due to friction from your feet or shoes in contact with the stairs, some is lost to air resistance and sound. ...

0

The wording of question is very confusing. If the object “travels back and forth” then it is already accelerating. Any additional force $F$ applied to the object, no matter how small, will change its acceleration by an amount $F/m$, where $F$ is the mass of the object. If the force has a vertical component and the object moves vertically then the force does ...

0

To give and intuitive idea: Imagine a person carrying a heavy suitcase fitted with wheels. As long as the person pulls his luggage on a flat terrain, he does (ideally) no work: The (gravitational) force is perpendicular to the motion. If the person arrives at a place when he/she has to pull his luggage say along an uphill ramp, then the more inclined the ...

0

It is not unlike dropping a mass in a constant gravity field. The force applied is constant (ideally) and the work is $F.dx$ in module (in the electrical example as well as in the gravitational example). In both cases, the object has lost potenttial energy and gained kinetic energy and the overall balance remains $0$.

-1

$$W=\mathbf{F} \cdot \mathbf{S}$$ (work done is the dot product between the force applied and the displacement of the system) There is another concept which is torque $$\boldsymbol{\tau} = \mathbf{F} \times \mathbf{S}$$ (torque is the cross product or vector product of force applied and displacement of the system) Dot product between the two vectors (both ...

0

work done in moving a charge across a potential difference = (𝐶𝑉)𝑉 = 𝐶𝑉2 This calculation is not correct. It should be $dW= (C \ dV) V$. Then $$W = \int_{v_0}^{v_f} CV \ dV$$ which for $v_0=0$ and $v_f=V$ gives the $\frac{1}{2}CV^2$ expression So the work done on the capacitor is equal to the energy stored in the capacitor, as must be the case for ...

2

In general, work is a dot product between two vectors. You must integate over the path, $W=\int \vec F\cdot \mathrm d\vec s$, but if the force is constant, this simplifies to: $$W=\vec F\cdot \vec s.$$ A dot product is mathematically the parallel components multiplied together. We could write it as $$W=\vec F\cdot \vec s \quad\Leftrightarrow\quad W=F_\... 0 Work done is a scalar quantity and it is defined by the product of two vectors Force and displacement. Now dot product of two vectors give a scalar quantity. So work done must be a dot product of the force and displacement vectors. That's why W=FS\cos(\theta) 1 Because the force applied on the object produces two effects - changing kinetic energy of the object by changing its speed and curving its path. The work is supposed to tell us, how much kinetic energy was transferred to the object (or from it). The kinetic energy is however increased only by the component of the force in the direction of the motion, which ... 0 This seems completely absurd, have I made an error in my experiment/logic? You have not made an error, the result that you have found is indeed correct and the only mistake is the mistaken feeling that it is absurd. Indeed, it is correct that the same \Delta v corresponds to different \Delta KE in different frames. The reason that this most likely feels ... 0 I think in this case it is necessary to do a closer examination of the concept of kinetic energy. Kinetic energy attributed to an object is kinetic energy with respect to some specified coordinate system. The kinetic energy attributed to an object is the amount of work that must be done to bring the object to a standstill. Let's say we have an object, ... 3 When talking about work, you have to specify the force that is doing the work: forces do work. The general expression for the work W done by a force \mathbf F over some path C is defined as an integral:$$W=\int_C\mathbf F\cdot\text d\mathbf x$$However, in the case of a force with constant magnitude F that is tangent to the path at all times, we can ... 2 Work is defined as$$W=\int\vec{F}\cdot d\vec{r},$$which is a line integral, so it depends on the path in general. However, if the force \vec{F} is conservative (\vec{F}=-\vec\nabla V), the work will only depend on the initial and final points W=V(\vec r_0)-V(\vec r_f). Thus, there are several situations that make the work be zero: The net force is ... 6 Gravity is doing work on you when you fall, even at terminal velocity. But, as you also point out, air drag does work as well. They are equal but with opposite signs: work by gravity is positive (trying to add to your kinetic energy), whereas work from air drag is negative (absorbing and removing energy from you, losing it as heat and movement of air ... 3 You are correct: zero total work is done on the object at terminal velocity. You could also ask "how much work is done on the falling object by gravity" and that would be nonzero. Likewise, you could ask "how much work is done on the object by the force of air resistance" and that would be nonzero. But, the total work done on the object ... 0 It turns out that terms representing the kinds of energy in a system generally are products of two factors. One factor is an extensive variable (depends on how much of the system you consider), and the other is intensive (does not). For example, PV: P is intensive, V extensive, or \mu N: \mu (the chemical potential of a species) is intensive, N... 0 What is the differentiating factor between work and heat?' Both heat and work are means of energy transfer. The main differentiating factor is heat is energy transfer driven solely by temperature difference whereas work is energy transfer driven by force acting through a distance. For example, The way in which entropy is defined is a very good way to ... 1 You are correct: heat transfer can be thought of as work done at a microscopic level. Consider the simplest example of two molecules colliding. There will be some electrostatic repulsive force for the very short duration of time when they are colliding, and this force will act over a very small distance. As a result, one molecule will gain energy and the ... 0 I think that we could take the definition for a hyperelastic material. From Wikipedia A hyperelastic or Green elastic material is a type of constitutive model for ideally elastic material for which the stress–strain relationship derives from a strain energy density function. Then, you will have something as the following$$S_{ij} = \frac{\partial U}{\...

3

The argument is probably this: it is assumed that the rubber body returns to the initial state; then its final energy has to be the same as the initial energy because internal energy of a body is function of state; same state means same energy. It is also assumed that no work was lost on heating the environment during this exercise. So if energy of both ...

1

the answer is indeed zero. also, you don't even need the integral because the force is constant. so you can simply do $W=\mathbf{\vec F}\cdot \Delta \bf \vec r$ where $\Delta \bf \vec r$ is your displacement.

1

It means that traveling along a conservative force increases your kinetic energy at the expense of potential energy, and traveling against it decreases your kinetic energy and adding it to the potential energy. The significance is in the negative work because a conservative force isn't associated with dissipative forces that waste energy. If it does negative ...

1

Stair machines are not exactly equivalent to stairs for a few reasons, but not due to the fact that you stay at the same height. In real stairs, the impact of your foot with the stair will be quite jarring (you'll use extra muscles for stabilization). The accelerations at the start and end of the step will tend to be different than the mechanics of an ...

0

find the....electric potential energy at the point the charge will reach after the known amount of work? Knowing the work $W$ in Joules that was done to move charge $Q$ gives you the change in electrical potential energy in moving the charge. It does not, however, tell you what the value of the electrical potential energy is after moving the charge because ...

-1

If $PT^2$ is constant, then, from the ideal gas law $P(PV)^2$ is constant. So $P^3V^2$ is constant, or, equivalently, $PV^{2/3}$ is constant.

0

you are confusing reference point and reference frame, in a particular frame of reference any point taken as the reference point in that frame, will have the total work done same but as the frame is accelerating we have to consider the work done by pseudo force as well.

1

By reference point, they don't mean reference frame. The proof is based around keeping track of the motion of one point on the body. Using the rotation matrix $[A(t)]$, based on that one point, you can determine the motion of any other point on the body. This reference point does not need to be anything specific, since for any point you choose, there is a ...

2

read that "Work done by magnetic field is always zero" This is misleading / not true in general. In macroscopic physics, (macroscopic) magnetic forces do work, for example magnetic force due to external field acting on current-carrying conductor $BIL$ does work when the conductor moves (such as in your example) and also this force can do work on ...

0

All the above answers are referring to classical mechanics of work which is defined as $W=F*d$, i.e. as linear momentum in a closed system in units of poundes*inches, but since there is no motion work equals zero. By Newton's conservation laws energy is conserved. If we look at system setup as an open system (Ludwig von Bertanlaffy's Open Systems Theory, the ...

1

As is often the case, a work/energy approach is the simplest method here. Can we assume that the nail is smooth, so no energy is lost due to friction between the rope and the nail ? If that is so, all we need to know to solve the problem is the height of the nail off the ground. By how far has the centre of mass of the rope been raised once it is hanging ...

2

When you move the object, it must have some speed. So in this case you have to accelerate the object and decelerate to get 0 speed at the endpoint. Thus when you calculate work done by a change in kinetic energy it's zero showing that you apply work W to accelerate and -W to decelerate the object.

4

There was a work $w$ to accelerate the object to a given velocity, and another work $-w$ to decelerate it to zero. The net work is zero. Remember that $dw = \mathbf {F.dx}$. When F changes direction the dot product changes sign.

3

First, there are two key statements that are important to keep separate. The work-energy theorem says that the net work is equal to the change in kinetic energy. The work done by a gravitational field, is equal to the minus the change in the gravitational-potential energy. The way I interpret your question, is that you move a block on a horizontal table ...

2

The curl of an electric field is given by the Maxwell-Faraday Equation: $$\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$$ When there is no time varying magnetic field, then the right hand side of the above equation is 0, and the curl of the electric field is just 0. When the curl of any vector field, say $\vec{F}$, is identically 0, we say ...

0

The fact that the vectorfield is conservative and has zero curl follow from the fact that it's the gradient of a potential. Since if $\mathbf{E} = \nabla \phi$ then $$\int_{\mathbf{r}_1}^{\mathbf{r}_2} \nabla \phi\, \mathrm{d}\mathbf{x} = \phi(\mathbf{r}_2) - \phi(\mathbf{r}_1)$$ so if the curve is closed then the work done is zero. The curl is also zero ...

1

You will have done work. Mathematically, $W = F \cdot dS$. Since that "particular point" moved, there is a nonzero displacement in the direction of the force, and work is done.

0

what you are thinking is partially correct. In simple terms, work done= force(dot product)displacement. So treating both as vectors, in case 1:- vector of force i.e. mg points downwards which is evident as gravity acts downwards and the displacement vector also acts downwards, now you can explain this with your own thought process that displacement is D(...

0

It's the particle that does work beacuse it's subjected to a force by the electric field. This force is $\vec{F}=q\vec{E}$ and work done by the particle from the point $a$ to the point $b$ is by definition: $$W_{ab}=\int_a^b\vec{F}\cdot d\vec{l}=q\int_a^b\vec{E}\cdot d\vec{l}$$ Remember that the electron has a negative charge. It does not make sense to say ...

1

I have learnt that at constant pressure, enthalpy change $∆H=q$ provided no non-expansion work is done. What you refer to as "expansion work" is normally referred to as boundary (expansion or compression) work, or $pdV$ work, which is applicable to closed systems (no mass transfer involved). The other possible type of work (non expansion work) is ...

4

For chemical reactions, the change in enthalpy $H$ (enthalpy of reaction) is, as you already said, defined as the heat $Q$ of reaction (assuming pressure $p$ is constant). This leads to $$\Delta_rH=\Delta U+p\cdot V\tag{1}$$ (the index $_r$ is for "reaction") This equation might seem a bit random at first, but it does make sense: We know that the ...

8

First, let's cover how a pushup is done on the floor. Your hand applies a force onto the floor, and the floor applies an equal and opposite force on you, which you use to push yourself up. The important part is that you push on the floor. However, the floor, being very rigid and massive, does not deform or move. Next, let's consider a trampoline or a ...

3

Conservation of energy. She starts off with the same kinetic energy $\mathrm{KE}_i$, and the work done by gravity is the same in each case, $W_\text{gravity}=mgh$, since each rock falls from the same height. Thus in each case, the same amount of gravitational potential energy is converted to kinetic energy, so the final kinetic energy is \$\mathrm{KE}_f=\...

0

I'm surprised that none of the answers here have used the easiest and most reliable solution using the electromagnetic field tensor. Consider a large wire oriented along X axis, carrying a current I in a static magnetic field along negative z axis. This will cause a force on the wire in upward direction and in rest frame and the work done is exactly the ...

4

You are doing exactly no work. Think of a table. With its normal force it can hold up an apple forever. Does it constantly spend energy on doing that? No. Otherwise it would eventually "run out", which doesn't happen. No energy is transferred between objects if no displacement or thermal interaction happens.

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