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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 ...

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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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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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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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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, $$N-mg=ma\text{, } ... 2 If \Delta S = r cos\theta then dl=ds=rd\theta and F_g=mg$$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 ... 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. 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 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 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. 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 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 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 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$$1 I suppose that one way to look at energy is that it is a convenient tool for book-keeping tool since it is a conserved quantity. It's actually amazing that there is something that we call "energy" which can exist in so many different forms (e.g., kinetic energy, gravitational potential energy, electrostatic energy, electromagnetic energy, etc.) but that if ... 1 In the first part you wrote E_{k}=E_{g} because kinetic energy is fully converted into potential energy. But in the second part, some of the initial kinetic energy (E_{f}) lost due to friction and part of energy left is E_{k}-E_{f} . Only this part is converted to potential energy E_{g} . Thus, E_{g}=E_{k}-E_{f} and this simplified as ... 1$$W=\int_{x1}^{x2}ma(s)ds$$Since ds/dt=v\Leftrightarrow ds=vdt:$$W=\int_{t1}^{t2}ma(t)vdt$$The expression for acceleration should of course be a(t), since it might depend on t (just as it depends on s). But writing it as just a is not "wrong" (the (t) just emphasizes the dependency on t, and a=a(t) still counts). To keep it consistent ... 1 You can describe the electric force it terms of potential energy, because it is a conservative force. In doing so you actually replace the concept of work done by this force by the concept of potential energy. So you can not longer use both descriptions simultaneously. If you describe the electric force as doing work, then you made positive work and the ... 1 You are in your reasoning overlooking something. Look at the diagram below: -q_1,+q_2 are two point charges at distance r. Coulomb's Law dictates that the attractive electrostatic attraction force between them is:$$F=k_e\frac{|q_1q_2|}{r^2}$$And the electrostatic potential U(r):$$dU(r)=F(r)drU(r)=-k_e\frac{|q_1q_2|}{r}$$Assume now that ... 1 The one with the greater magnitude or absolute value is greater. The sign has no bearing. Negative work only tells us about the direction in which the work is being done, positive along the direction of motion, and negative anti-parallel to the direction of motion. An object in a gravitational field can escape to infinity if TE>0. GKE has to be greater ... 1 I asked a somewhat different, yet similar question.Hope this helps! Why is an LC oscillator lossless, but C V^2 / 2 energy is lost to a capacitor connected to an ideal voltage source? 1 There are different forms of energy. Energy can be converted from one form to another but cannot be destroyed. In this case the kinetic energy of the hammer is driving the nail into the wood which is breaking the molecular bonds in the wood fiber. The energy is converted to heat energy as a result of the breaking of the bonds and the friction of the nail in ... 1 The Newton unit is not a fundamental unit but consists of:$$\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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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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