# Tag Info

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The significance is that it is the work done by the fluid. Moreover, $$W=\int_{V_i}^{V_f} P\,dV$$ where $W$ denotes the work done by the system during the whole of the reversible process. This relation appears regularly in thermodynamics, usually as the first law of thermodynamics $$dU=\delta Q+\delta W=\delta Q-pdV$$ (the minus sign represents ...

0

Work is the change in kinetic energy. In both cases, the man starts on the ground at rest and end on the chair at rest. In both cases, the net work is zero. If you want to be more specific and describe the work due to his muscles, that positive work must exactly offset the negative work due to other sources. Gravity does negative work equal to minus the ...

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I'll ask the question for completeness. As discussed in the comments, if you think about all the posible ways of dissipating energy, maybe you will doubt of the veracity of the assertion made in the exercice. But, the clue is not to think in if the teacher wanted to be "captious"... but that in this case all these ways of dissipating are really small ...

2

If $\Delta S = r cos\theta$ then $dl=ds=rd\theta$ and $F_g=mg$ $$W_{g}=\int_0^\pi mgrcos\theta d\theta$$ If you're taking the angle from the center of the circle (which you are, since you said that $\Delta S = r cos\theta$, then the initial position of the ball is $-R$, since displacement is a vector quantity (and the final ...

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I asked a somewhat different, yet similar question.Hope this helps! Why is an $LC$ oscillator lossless, but $C V^2 / 2$ energy is lost to a capacitor connected to an ideal voltage source?

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The one with the greater magnitude or absolute value is greater. The sign has no bearing. Negative work only tells us about the direction in which the work is being done, positive along the direction of motion, and negative anti-parallel to the direction of motion. An object in a gravitational field can escape to infinity if TE>0. GKE has to be greater ...

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Energy can neither be created and nor be destoryed it can changes from one form to another form. For daily life example just when we do any work as for example just when we play a football and kick in the football and take our energ to football to kick it then energy is going into football and it can move from KINETIC ENERGY INTO POTENTIAL ENERGY

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Simply: The magnetic field does no work on a charged particle, because it's always normal to the particle trajectory. Work equals the change in kinetic energy...i.e. no change in speed.

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You are in your reasoning overlooking something. Look at the diagram below: $-q_1,+q_2$ are two point charges at distance $r$. Coulomb's Law dictates that the attractive electrostatic attraction force between them is: $$F=k_e\frac{|q_1q_2|}{r^2}$$ And the electrostatic potential $U(r)$: $$dU(r)=F(r)dr$$ $$U(r)=-k_e\frac{|q_1q_2|}{r}$$ Assume now that ...

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You can describe the electric force it terms of potential energy, because it is a conservative force. In doing so you actually replace the concept of work done by this force by the concept of potential energy. So you can not longer use both descriptions simultaneously. If you describe the electric force as doing work, then you made positive work and the ...

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1.) You lose energy by doing work. The difintion of work by itself is the product of force and dislplacement(if the force is onstant) or else $\int_{}{}F dx$ for non constant forces.In case the case of doing a plank, you do zero work though you are resisting the force of gravity. So in terms of physics, you lose 0 energy even though you get tired. The reason ...

2

When you compute the work of individual forces it does not matter what causes the displacement. As long as the force is acting while the object is being displaced, it has the potential to do work. Both gravity and the normal on the head are perpendicular to the displacement and do not perform work. Friction (a force made by the porter over the luggage) is ...

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The weight of the luggage is in downward direction. So to balance the gravitational force of the luggage, the porter have to apply the force in upward direction. That's why the angle between the force applied and the displacement is 90 degree. So the work is 0. So in simple words, its not really the gravitational force but the force applied by us on the ...

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Work and kinetic energy are interchangeable, so they are the same, ie. they have the same units. Your work can give kinetic energy to a body and a body with kinetic energy can produce work.

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The Newton unit is not a fundamental unit but consists of: $$\mathrm{[N]=[kg\cdot m/s^2]}$$ which you can convince yourself of from Newton's second law $F=ma$. Plug it into $\mathrm{[N\cdot m]}$ and you'll see.

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I did not quite understand your problem in solving a simple question. But i think i get your doubt. The friction when ever taken in the work energy theorem is always taken as negative and as the work is done against the direction of the gravity the work is negative. here do not worry about the potential energy just calculate the work done when you equate ...

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As the total work on the system equals the change in kinetic energy: $$W_{Friction}+W_{Potential} = \Delta K$$ Taking: $$v_{i}=v_{f}=0$$ we can write $$W_{Friction}+W_{Potential} = 0$$ $$W_{Friction}+\Delta V = 0$$ $$W_{Friction}-V_{i} + V_{f} = 0$$ $$W_{Friction} + V_{f}= V_{i}$$ $$E_{i} = W_{Friction} + E_{f}$$ obs: Work by definition is $$\int ... 0 Yes and no, but I think in your context the answer is yes: only account for friction work. This is a good place to consider that energy accounting depends on the choice of system. In introductory courses, potential energy is usually introduced as gravitational potential energy, mgh. IMO, this is dangerous because it can lead to confusion, as we see in ... 0 When you have a non conserving force in the system(such as friction) an energy only approach only gives you an estimate of the results. It's not a particularly good way to look at the situation. The initial energy is going to be the kinetic energy. The resulting energy will be the gravitational potential less the friction losses. The frictional losses ... 1 There are different forms of energy. Energy can be converted from one form to another but cannot be destroyed. In this case the kinetic energy of the hammer is driving the nail into the wood which is breaking the molecular bonds in the wood fiber. The energy is converted to heat energy as a result of the breaking of the bonds and the friction of the nail in ... 3 That's not quite how the brain works. Not a lot of mass actually moves, but the electrical impulses with which neurons communicate require the repetitive movement of electrical charge against differences in electrical potential, and that takes work. In fact, the human brain requires significant energy to do its job. Quoting from Appraising the brain's ... 0 HINT: WORK-ENERGY PRINCIPLE The work done by all forces acting on a particle (the work of the resultant force) equals the change in the kinetic energy of the particle. Here the answer will be option-3 equal work done. 1 Yes you can. You have to "walk" in such a way that you will exert a force that is perpendicular to the motion of the escalator. Of course that means you will accelerate toward the bottom of the elevator, at the same rate as if you were on a frictionless slope. You can still accend the elevator if your initial speed is high enough. (Please don't try ... 1 Because the work done by friction is converted into rotational kinetic energy of the cylinder, since friction provides the torque to roll down the cylinder. 1 It is more difficult because it requires more energy. Once you are up to speed, walking on a flat surface requires only enough energy to overcome friction, air resistance, and the energy to move your legs. Walking up stairs is more difficult because you additionally have to provide the energy to lift your body weight up the stairs. It's the difference ... 1 When moving up you are pushing yourself in the opposite direction of the force of gravity. Therefore you do a positive work which is approximately mgh (h is the height of the step). While coming down gravity will do the same work for you. 2 You have to lift up your body mass under the presence of gravity, so you have to overcome the force of gravity. While climbing up the stairs you have to put force on ground, more then your weight, which put same force but in opposite direction i.e. on you. Suppose your mass is m, then while climbing up, if you resolve the forces then,$$N-mg=ma\text{, } ...

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case Ist: work done by object On object there are four forces- $mg$, normal force, force exerted by man and frictional force. Normal contact force and $mg$ are perpendicular to the direction of the force exerted by the man and the frictional force hence not going to include them but in x-direction force by man and the friction are opposite hence the net ...

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The work done by the person equals the work done on the object by the person, but it is not equal to total work done on the object, because friction forces do work on it as well.

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The answer is actually no. The kinetic energy of the object is not changing, hence the net work on the object is equal to 0 J. You are doing positive work on the object, but friction is doing negative work; taking energy. So you can still be doing work on the object, but the net work on the object, which is what changes the kinetic energy, can be 0. This is ...

0

Note that the sign of the expression $$\frac12 \;k \; (x_f^2 - x_i^2)$$ depends on your choice of coordinate system. If you put $x_f=0$, then the result is negative; if you put $x_i=0$ then the result is positive. But the work done should be the same, regardless of the choice of coordinates. So what is happening? Simply, when we say $F = -kx$ we have ...

1

$$W=\int_{x1}^{x2}ma(s)ds$$ Since $ds/dt=v\Leftrightarrow ds=vdt$: $$W=\int_{t1}^{t2}ma(t)vdt$$ The expression for acceleration should of course be $a(t)$, since it might depend on $t$ (just as it depends on $s$). But writing it as just $a$ is not "wrong" (the $(t)$ just emphasizes the dependency on $t$, and $a=a(t)$ still counts). To keep it consistent ...

0

This question arises because of a subtlety involved in the choice of variables and infinitesimals we use in the definite integration to find the work done by the force. If the block is at the coordinate $x$ and is moving towards the origin then the work done by the spring on it during the small interval of time in which the block travels a distance of $dl$ ...

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Notice, we know $$x_f<x_i$$ $$\iff x_f^2<x_i^2\ \ \ \ \ ( \forall \ \ \ x_i, \ x_f\ge0)$$ $$\iff x_f^2-x_i^2<0$$$$\iff k(x_f^2-x_i^2)<0 \ \ \ \ \ ( \forall \ \ \ k>0)$$ Now, the work done by the spring on the block is $$\int_{x_i}^{x_f}kx\ dx=\frac{1}{2}k(x_f^2-x_i^2)<0$$ Thus, the work done by the spring on the block is negative. It ...

1

In the first part you wrote $E_{k}=E_{g}$ because kinetic energy is fully converted into potential energy. But in the second part, some of the initial kinetic energy $(E_{f})$ lost due to friction and part of energy left is $E_{k}-E_{f}$ . Only this part is converted to potential energy $E_{g}$ . Thus, $E_{g}=E_{k}-E_{f}$ and this simplified as ...

1

I suppose that one way to look at energy is that it is a convenient tool for book-keeping tool since it is a conserved quantity. It's actually amazing that there is something that we call "energy" which can exist in so many different forms (e.g., kinetic energy, gravitational potential energy, electrostatic energy, electromagnetic energy, etc.) but that if ...

1

The initial kinetic energy $E_k$ gets partly dissipated as friction, $E_f$, and partly converted to gravitational potential energy, $E_g$. The sum of these two must equal the original energy input, so $$E_k = E_f + E_g$$

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One thing that may help you is to understand that work is an expenditure of energy. For instance, if you push a rock up a hill, you exerted a force over a distance, so you had to use energy. But the coat hanger in your closet isn't expending energy, even though it's exerting a force to hold itself up over a period of time - because it isn't moving over any ...

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First of all, forces accelerate an object when the net force is not zero. If friction is present, it actually does accelerate an object. Acceleration is defined as the change of velocity over time. This is not limited to increasing speed. Gravity also can accelerate an object. Work is just defined as $Fd$ (force times distance) just because it is useful. ...

2

This is a bizarre question. Newton's laws do include internal forces. However, Newton's third law happens to cancel out their overall effect on a center of mass. But, if you want to understand the motions of the constituent parts of the system, then you do have to understand their internal forces. So let's assume that we have a collection of particles ...

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The idea of net work encompasses the fact that while you may have to accelerate the mass initially to get it to move, it will also decelerate later (in other words, you can stop applying force a little bit before you reach the end, and let its momentum carry it the rest of the way). Let's show this with a simple example. Assume I am trying to lift a mass M ...

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This is really just a simplified version of Timaeus' answer, so please accept his answer not mine. Anyhow, you're quite correct that the ball gains energy, but that energy doens't appear from nowhere. in any (inertial) frame total energy is always conserved. What you are seeing is some of the kinetic energy of me and the room being transferred to the ball. ...

1

Try arresting the balls motion while standing on a path of ice. What happens is the ball and that floor come to a relative rest. And any force exerted on the ball to slow it down or speed it up has an equal and opposite force exerted on the thing accelerating it. You can analyze it in any frame and get a correct description (though kinetic energy depends ...

1

The question, as written, has no need for calculus. Find the volume of water needed to fill the frustum. http://jwilson.coe.uga.edu/emt725/Frustum/Frustum.cone.html Find the weight of that water Find the work done to pump all that weight up 6 feet over the top of the wall of the tank, letting it splash down into the tank. Done.

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We know that the gravitational potential energy of an object is $mgh$. Initially, $h=0$ for the water before it is pumped in, so the potential energy is 0 as well. From here, this is a related rates problem, the rates you want to be relating is the rate of increase of the gravitational potential energy of a horizontal layer of the water filling the tank to ...

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