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From what I remember $h=\frac{2r}{3}$ is indeed the corect answer. This the incorrect part of your reasoning. The component of the mass's weight along the centre disappears only when θ >becomes 90 degrees. At this point, it leaves the surface of the hemisphere. From what I understand your saying that at $θ = 90 degrees$ the radial component of the ...

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How do voltage and voltage drops over a circuit relate to work done? The Volt unit is energy normalized to unit charge; Joule per Coulomb. Since the Amp unit is Coulomb per second, the product of the voltage across and current through a circuit element is the power associated with the circuit element. For a DC circuit, voltage and current are constant ...

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But then that means that the electron in the 5 ohm circuit would have done 5x the amount of work (or work done on it) of the 1 ohm circuit over 5x the duration. You're confusing work with power here. Work has nothing to do with duration. If an electron crosses a potential difference of $V$ with any resistance in between, the work done is the same, ...

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Actually I think I disagree with the answer by BMS (the group of asymptotic symmetries of asymptotically flat spacetimes?). However I am not sure to have understood BMS'answer completely. In my opinion, there is no difference between the definition of work in pure mechanics and work in thermodynamics (I stress that I am speaking of thermodynamics and not ...

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Here is one issue where thermodynamics and mechanics could differ in the definitions of work. In mechanics, a non-careful, ambiguous, but common definition for the work done by a force $\vec{F}$ is $\int\vec{F}\cdot d\vec{s}$. The problem with this is that we're not told which infinitesimal displacement $d\vec{s}$ to use; one could use (1) the infinitesimal ...

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As Emilio Pisanty asserts, the magnetic Lorentz force $\boldsymbol{F_m}= q \boldsymbol{v \times B }$ does not do work, since $\boldsymbol{F_m \cdot v} = 0$. I don't have access to Griffith, but I find Feynman's discussion (in v. II, ch 15) confusing, since he has magnetic forces doing work on the current while also asserting those same forces do no work. ...

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Magnetic fields never do work directly. This is because the magnetic force on any charged particle, $$\mathbf{F}=q\mathbf{v}\times \mathbf{B,}$$ is always orthogonal to the velocity, and therefore the power transferred, $\mathbf{F}\cdot\mathbf{v}$, is zero. On the other hand, this seems to contradict much of our intuition about how magnets behave. If you ...

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You have to do some work on the object to keep it moving at a constant speed! Friction against the floor would stop it dissipating its kinetic energy as heat. You need to push the object applying to it a net constant force in order to keep it moving, so you do some work equals to the net force you do on the object times the path you make the body take. That ...

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Let's say you roll a ball (of mass $m$) down an inclined plane of angle of inclination $\theta$ and coefficient of static friction $\mu_{static}$. Then you know a force parallel to the inclined plane acts on the ball through its center of mass. Another force parallel to the surface acts in the opposite direction of motion as follows, The force $\vec F = ... 2 When a disk or other object is rotating on a horizontal surface with constant velocity, there is no static frictional force. Your logic is correct: if there were a horizontal force, the center of mass would be accelerating. If the rolling object suddenly encounters a frictionless surface, it would continue to satisfy the rotating without slipping condition. ... 0 The work done by a force along a path$\gamma\$ is defined as $$W = \int_\gamma \vec{F}(\vec{r})\cdot d\vec{r} = \int_a^b \vec{F}(\gamma(t))\cdot\dot{\gamma}(t)\,dt$$ where the last equality is the actual definition of the line-integral of a vectorfield along a curve. Note that the force is explicitly depending on position. As you state correctly, a force ...

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Does the work done by the force remain 0 even if it varies at all points on the loop ? Yes. For example, the gravitational force. Note, that in general fields are not conservative. So if you write an arbitrary force, the work will not be zero.

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You are on the right track for the minimum energy required. It has to be at least that required to lift the weight over the height change. However, there are so many other things envolved that this minimum energy per squat won't be a very useful figure, although the legs acting a levers has nothing to do with it. Either way, the body weight is raised, ...

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