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0

The force is perpendicular to the motion of the charge carriers. But the resulting motion of the wire is in the direction of the induced force, so work is done on the wire.


0

Lots of good answers here, but most of them are pretty mathematical and not very intuitive. Lets consider a realistic example. You're on the moon with a six shooter and some extra bullets. You are in a uniform field, and you make two point masses travel through the same distance by dropping a bullet with one hand and firing at the lunar surface with the ...


1

Work done is also defined as change in kinetic energy of the body. Since F is constant force so F/m=a is a constant acceleration of m. So, $$v^2-u^2=2ad$$or$$mv^2/2-mu^2/2=2mad/2$$which is the work done by the force. The body has travelled d distance with accleration a in the force field assuming u was a constant velocity when it entered the field and v is ...


1

Without any math and considering only Newtonian model here, I would say that if you move the inertial system at the same speed and direction as your mass point is moving, than you have no initial movement of the mass point and the total force used for acceleration will be the same as if you calculated or measured it in the original inertial system.


6

Well, you simply need to accept that work is given by Force time Distance, and it doesn't matter how long it takes. For example, the work done on a mass $m$ lifted a distance $h$ against gravity with an acceleration $g$ is given by:$$W=F\times h=mgh$$ If you are told that someone is going to drop a $1$ kilogram mass on your head from a height of $10$ ...


4

Well, the reason it doesn't matter is that work is defined as $$W = \int\vec{F}\cdot\mathrm{d}\vec{s}$$ so if you keep the force the same and the distance the same, this remains the same, regardless of what you do with the initial velocity. Of course, that definition probably isn't particularly satisfying. So consider this: when an object is subject to a ...


2

As you describe, the definition of work is just: $W=F d$. What you are confusing maybe is the rate of work $P$ and the force $F$. When you move fast, $P=Fv$ is larger, however the travelling time is shorter. let's consider we are moving in a constant velocity. Then: $$W=Pt=Fvt=Fd$$ Independent of velocity.


2

As you note, for a constant force acting on an object which moves in one direction, the work done is equal to $Fd$. One can see from the equation that work is not dependent on time, but only on force and displacement. In order to conceptualize this, you could think about the energy involved in the situation you describe. When work is applied by an external ...


0

The OP and Comments give two opposite answers as correct for this question Here's the one I think is correct: There is a system, consisting of two blocks connected by a rope. It is initially at rest, and after a period of time, it (all of its parts) is moving. It has gained kinetic energy, and therefore work has been done on it by an external force. Both ...


0

You can ask for the work done by an individual force, or the total (or net) work done on an object. Consider a mass attached to a spring, the mass initially at rest. I apply a constant force of 1 N, and the mass moves 1 m and then stops. The work done by my hand on the book is 1 J. The force and displacement are in the same direction, so the work done by ...


0

Usually problems are framed with "before" and "after" states. A book was here, a force applied, and later the book was there. So there might be an implicit time interval, but that interval is not used in calculating the work. Also, frequently, the object starts at rest and ends at rest, also providing a time interval, which again is not used in ...


1

When I was in high school, my teacher would use a mouse and an elephant for this kind of problem. Let us put a mouse and an elephant on ice skates. You push the elephant, whose mass is $m_e$, with force $f$. The elephant slowly accelerates. You keep pushing until it eventually moves a distance $s$. By this time, the elephant is slowly moving at velocity ...


0

You have to realize that the two bodies will reach a different final speed. Both will have the same kinetic energy (equal to the work of the force), but the larger one will have a considerably smaller speed.


1

The force is only applied for that distance, the fact that it is the same implies that the smaller object will have lower speed. Note that the relation does not imply the distance the object will move; instead, the displacement is the distance over which the force is applied. $$Fs=\frac{1}{2}mv^2$$ Hence, because the objects have the same energy but the mass ...


0

I mean when do we say a force has worked on a body? Whenever a force($F$) acts on a body and creates a displacement($S$), we say work has been done by that force on the body. Yes that applied force can work against another force as well, but the point to note that THERE IS A FORCE ON THE BODY. The basic equation is : $W$=$FScos\theta$, where $\theta$ is ...


1

It is the amount of energy $E$ that must be transferred to an object through Force so that it can get displaced by some distance $s$. In completely layman's terms, the amount of energy someone must spend to do something.


3

The integral of a scalar function / vector field along a curve is defined with reference to a parametrization, e.g for a scalar $$ \int_C f \mathrm{d}s := \int_a^bf(r(t))\lvert r'(t)\rvert \mathrm{d}t$$ where $ r: (a,b)\rightarrow \mathbb{R}^n $ is a parametrization of the curve $C$. It is then shown, that the integral is invariant under reparametrization. ...



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