# Tag Info

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No body is perfectly rigid. If you want to consider friction as due to small bonding sites, then those sites do move (strain) under load (stress). So the premise that no work is done because the binding locations are fixed is incorrect.

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in which the fact that "(...it's certainly true that the horse is exerting a force F that is greater than the force on the train)" F is greater than the net force $F>F\cos{θ}=F_{net}$ is attributed to the way muscles work. No, it's not. This is just a property of the dot product. The comment on muscle efficiency was merely an aside which, in ...

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You didn't do anything wrong, just incomplete. As @Sebastian alluded to, the work will be $$W=E_{3R}-E_{2R}=T_{3R}+V_{3R}-T_{2R}-V_{2R}$$ where the subscript denotes orbital position. You already found the $V_{3R}-V_{2R}$ term, now just use the centripetal motion formula $$\frac{mv^2}{r}=\frac{\gamma Mm}{r^2}$$ to find the difference in kinetic energy ...

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No. The easiest way to see this without invoking rotating reference frames is to write out Newton's Law's in polar coordinates, which work out to be: \begin{align*} F_r &= m \ddot{r} - m r \dot{\phi}^2 \\ F_\phi &= m r \ddot{\phi} + 2 m \dot{r} \dot{\phi} \end{align*} From these, it's pretty easy to see that if we have $F_\phi = 0$ and $\dot{r} ... 0 In a rotating frame of reference you usually consider the centrifugal force, one of the three possible fictitious forces. This is because you usually consider it as rotating at constant angular speed. There are other forces however: the Coriolis force that appear when the object moves in the rotating frame of reference. The Euler force when the angular ... 0 No. W=o for internal forces is valid for all bodies irrespective of their rigidity. Because a non rigid body can b deformed only due to external forces. Not due to internal forces. Because internal forces acts along the same line. With equal and opposite direction causes net force to b 0 0 Displacement refers to the object's position relative to the observer. The "place in space" of the orange. Distance is the object's position relative to an earlier position. If you pick up an orange, and run 10 miles holding it straight out, no work gets done on the orange. But if you extended your arm 10 miles, you would have to be doing work on the ... 17 It depends on whether the force field is conservative or not. Example of a conservative force is gravity. Lifting, then lowering an object against gravity results in zero net work against gravity. Friction is non-conservative: the force is always in the direction opposite to the motion. Moving 10 m one way, you do work. Moving back 10 m, you do more work. ... 3 If you 'carry' an object 10 meters in one direction then return it back 10 meters from where you started the work done on the object is not the force you expended times distance walked. The formula you write is often misunderstood and misused. In your example, when you lift the object in a gravitational field, the work being done on the object is its weight ... 0 The answer depends on how particular you want to be about it. Upon first examination you might say that in the Earth's conservative gravitational field, that the work must be the same, since the start and endpoints are the same and we have ignored any non-conservative forces (e.g. friction, air resistance, etc.). This is more than likely the response that ... 1 The "work" is the useful rotational pull you get from the motor, and excludes any wasted heat. If only 75% of the energy going in comes out as work, then you need to put in 90/0.75J of energy to get 90J of work out. 1 Since$\vec{F}=m\vec{a}$, the force of the magnetic field causes the charged particle to accelerate. An acceleration, however, is a change in velocity, not necessarily a change in magnitude of the velocity. If the force is parallel to the velocity, the magnitude of the velocity will change. If the force is perpendicular, only the direction of the velocity ... 0 Maybe one intuitive explanation is that the change in velocity will be$O(\Delta t)$, while the change in speed during a little time$\Delta t$will be$O(\Delta t^2)$. In fact if the velocity$\mathbf v =v_x\hat x$is initially in the$x$and the acceleration in the$y$direction, then after$\Delta t$: $$\Delta v^2 \approx v_x^2+(a\Delta t)^2-v_x^2=a^2 ... 10 I know that magnetic force acts perpendicular to the direction of the original velocity No, the magnetic force acts perpendicular to the current velocity. Once the direction of the velocity changes, the direction of the force changes as well. Cast in math:$$ m\dot{\vec v} = \vec F_L = \frac q c \vec v \times \vec B $$From this we get (v = |\vec ... 1 I have highlighted some key word lacking in your revision. Also, work has a very specific definition. The difference in gravitational potential difference between \vec{r}_1 and \vec{r}_2 is the negative of the work done on a unit mass by the external gravitational field as the unit mass moves from \vec{r}_1 to \vec{r}_2. As an example, consider a ... 1 Words are imprecise. Your wording is much less imprecise than the one from the textbook. As the work is positive or negative (depending on the direction taken), there is no need to define a "positive direction", but the potential difference depends on the order of \vec r_1 and \vec r_2 (that is: if the potential difference moving from 1 to 2 is positive, ... 0 Work is defined by the line integral W=\int_\gamma F\cdot dr, where W is work, \gamma is some particular path, and F = F(x,y,z) is a vector field. Now let's consider a fact we know from calculus I:$$\int_a^b f(x)\,dx = -\int_b^a f(x)\,dx$$Likewise, when we take the path in reverse (I suppose we could call it -\gamma), we'd get$$\int_\gamma ... 0 The total work is simply the algebraic sum of the negative and positive work. 2 So the net work on the object is zero and it doesn't gain any energy. The object obviously did gain energy. The object's potential energy increased by$mgh$, and its kinetic energy didn't change. So what's going on? Is the work-energy principle wrong? The answer is no, the work-energy principle is not wrong. The work energy principle merely says that ... 0 Gravitational weight is a conservative force so work done against it is entirely equal to the change in potential between the start and end points (in the absence of dissipative forces like friction). The internal energy of the mass does not change as it is moved upwards and your assessment of the energy considerations is correct. Consider alternatively the ... -1 Gravity doesn't do -mgh work on the object while it is lifted, gravity converts the potential energy gained while lifting the object (mgh) into kinetic energy (1/2 mv^2) after it's dropped. 2 The total work done on the object is the change in kinetic energy:$W_{total} = \Delta E_K$* While the gravitational potential energy of the object is:$U_G = mgh$So, although it costs energy to lift the object up, the total work done on it is$0$because both at the beginning and at the end it has no kinetic energy ($v_{i,f}=0 \rightarrow E_{K_{i,f}}=0 ...

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We consider the integral: $$\sum_{i\lt j}\int_{t_0}^{t_f} \mathbf{F}_{ij}(\mathbf{r}_j(t))\cdot(\mathbf{r}'_j(t) - \mathbf{r}'_i(t))dt$$ For a rigid body, the distance between any two masses is always held constant, a fact that we can express as: $$\vert\mathbf{r}_i(t) - \mathbf{r}_j(t)\vert^2 = \Delta_{ij}$$ or (\mathbf{r}_i(t) - ...

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