# Tag Info

1

This is a question everyone asks at first because it intuitively seems like a contradiction. However, it is not. Conceptual Examples I think you are not far off but perhaps the third law is the one tripping you up, not the 1st... But anyway, here are some conceptual examples, which might help... Example 1. Consider the particle in the frame for a ...

21

In introductory problems about work you're normally taught that it's force times distance: $$W = F \times x$$ and you treat the force as constant. If you look at the problem this way then you're quite correct that if the force is $F = mg$ then the box can't accelerate so it can't move. However a more complete way to define the work is: $$W = ... 3 Newton's first law states that: "An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force." It is often called the law of inertia. So if You want to move an object with zero velocity, at first moment You have ... 1 external force acting on the box by us should be a little more than the weight, otherwise indeed: no acceleration! So the mgh is really a lower limit. We need to accelerate. And, by the time we reach h, decelerate -- so we get this extra little bit of work back. But only in a physics sense. 1 Work is described by the following formula:$$W=\vec{F} \cdot \vec{s} = F\cdot s \cdot \cos \alpha$$with s is the distance, 40 m in this case, and \alpha the angle between \vec{F} and \vec{s}, which in this case happens to be zero, giving W = F \cdot s F is force, in this case work is done against gravitational force, so$$F=mg$$m is the ... -1 Since the work done must be W=F\cdot d And we need to calculate just the distance, because we already have the force. x(t)=x_0+v_ot+\frac{1}{2}at^2=\frac{1}{2}at^2 Finally we reemplace that in the fist equation to find the work done. W=F\cdot\frac{1}{2}at^2  0 The friction acts only when the contact point slips relative to ground. You can consider speed of lowest point to be sum of v and \omega r with proper directions. Friction acts till there is slipping and condition for no slipping is v=\omega r when v is right and \omega is clockwise acc. to diagram As in first case, the lowest point in always ... 0 1st que- as the point in contact of the disk and the plane is not moving, no work is done. and as for the second question; no body is perfectly rigid, therefore due to deformation there is loss in energy. that is why the disk stops. 0 Power is, in words, the rate at which work is done and work done equals the amount of energy converted from one form to another. For electric circuits, the power associated with a circuit element is the product of the voltage across and the current through. We can verify this via dimensional analysis:$$v \cdot i = \frac{J}{C}\cdot \frac{C}{s} = ...

0

Starting from the definition of power $$P=\frac{dW}{dt}$$ We can solve for the work (after integrating). The thing to know is what power is in terms of current and voltage or resistance $$P=IV=I^2R=V^2/R$$ Clearly, $P=IV$ is what we want to use. Last thing is what is $V$ for a inductor? It is $V=L\frac{dI}{dt}$ From here I will let you (and future ...

1

If this were a mechanical pump, then the work done per unit volume pumped would be: $$W = VP$$ where $V$ is the volume pumped and $P$ is the pressure increase across the pump. According to Wikipedia the volume pumped per heartbeat (at 72 bpm - does it change with pulse rate?) is about $70$ cubic centimetres per beat, which is $7 \times 10^{-5}$ m$^3$. ...

0

Let me try to attempt to clarify in some direction that I feel comfortable. From usual mechanics, you know the definition of work is given by : $$W = \int \bf{F}.\bf{ds}$$ for simplicity considering the work done to be in the same direction as displacement $$W = \int {F}{ds}$$ Now this is just one version of defining work. In generalising, we redefine ...

0

It may just be the wording of your text that provides the confusion. Work is the same in thermodynamics as it is in mechanics. Work is an energy transfer process. In thermodynamics this often equates to energy in the forms of pressure, temperature, volume, etc. that is transformed into energy associated with position or movement (i.e. kinetic and potential ...

2

You may think you are not moving when you plank, but your body maintains plank by pushing you back up imperceptibly after you droop imperceptibly. Your muscles do work against gravity. I can't imagine how to estimate how much.

2

You have the right idea, but the question asks for situations where there is a force and no work. Centripetal force does do no work but there is a force, so I is true. In III you are exactly right, but it says there is a force and no work, which falls under the question. I think you have misunderstood the question. II is false because a force in the ...

0

You have the ideas down, I think, but for some reason, you seem to have not yet noticed the disagreement between your thoughts. For instance, if you've already stated that $W = fd$, and that you're looking for examples of measurable force with no work, what would that tell you about $f*d$? So if work is $f*d$, then in order to have work, you must have some ...

5

Your teacher's explanation is incorrect. A simple counterexample can be constructed to illustrate this by considering what happens when the role of your arm is replaced by that of a rubber band. When a weight is suspended from the ceiling by a rubber band, the band stretches and its polymer chains become more ordered, in exact analogy to your teachers ...

5

Think about the work-kinetic energy theorem, which states that the net work done on an object is equal to its change in kinetic energy: $$W_{net}=\Delta\mathrm{KE}.$$ You are right that when lifting an object of mass $m$ by a height $h,$ in a uniform gravitational field, the work you do is $W_{you}=mgh$ (assuming, as you said, that you're applying a force ...

3

The electrostatic potential energy of a system of point charges is defined as the work required to be done to bring the charges constituting the system to their respective locations from infinity. Suppose we have a configuration of point charges. If the potential of the energy of the system is negative, this means work is positive. Consider two point ...

1

With work you also have to take care in specifying what is doing the work. If it is the work required to move a particle or the work done by the electric field, these have different signs. If the work done by the electric field on a point charge is positive it means it is moving in the direction of the electric force acting on the point charge, therefore, $W ... 2 Rule of thumb for working it out: If you imagine letting a charge go, the direction it tends to move is toward lower potential energy. The opposite direction is toward higher potential energy. This is independent of the choice of where the zero of energy is. 1 Have to be a bit careful with potential energies, as the 0 point of potential can be arbitrarily chosen. Only changes in potential are well defined without a choice of 0 point. That said, it is often convenient to choose the 0 point at$\infty$, and this is the typical choice when talking about assembling point charges. With this choice, the potential ... 0 After some thinking, I got the answer. When you decrease the ambient pressure from$p_i$to$p_{atm}$, the system goes out of equilibrium, and so the expansion of the gas is not quasi-static. So, all the state variables like pressure and volume aren't defined. So we can't use them for calculating the work done in definition 2. 4 Good question, and the answer is that$W' > W$if$F > mg$. The reason for this is that if$F = mg$then the net force is zero so the particle travels at a constant velocity. That means it's kinetic energy hasn't changed so the only change is the potential energy. However if$F > mg\$ then the net force is positive and the particle is accelerating ...

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