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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. ...

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 ... 4 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 haveF_\phi = 0$and$\dot{r} ...

3

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) - ... 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 ... 2 Let's set up an example. Imagine a 10kg platform, a spring, and a 1kg mass. Suppose the spring is compressed to hold sufficient energy to push the items apart with a relative velocity of 11m/s. Beginning from rest, this will give the object a forward speed of 10m/s and the platform a rearward speed of 1m/s. The total kinetic energy after this ... 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 ... 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 ... 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, ... 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 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 ...

1

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 ...

1

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.

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 ...

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.

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