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I think the confusion arises due to difference in equations from Physics and Chemistry text books. The physics text writes $U=q-w$ while chemistry says $U=q+w$. From the physics text its mentioned that work done by the system is positive means $U=q-(+w)$ and work done on the system is negative means $U=q-(-w)$. THEREFORE, both are correct in their own ...

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In the following diagram, is work done by static friction 0 ?, since the point of application is also moving with speed v w.r.t. ground here and is only stationary w.r.t. the block on which sphere is rolling w.r.t. ground here. Static friction itself is 0. The formula $f_s=\mu N$ defines the maximum possible magnitude of the static friction force, not ...

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Officially, I completely agree with the other answer given. I would like to offer this answer as a simplistic, intuitive answer to the question. No math involved. I understand where your question comes from. In fact, depending on your current level of education, this question could indicate a high potential for future scientific success. We all know that ...

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In polar coordinates, the velocity being tangent to the circle, it is directed along the $\hat{e}_{\theta}$ vector. The centripetal force is directed along the $\hat{e}_r$ vector. So $\frac{F}{m}\hat{e}_r = \frac{d\vec{v}}{dt} = \frac{d(|v|\hat{e}_\theta)}{dt} \underbrace{=}_{\text{Chain Rule}} \frac{d|v|}{dt}\hat{e}_{\theta} + ... 5 The work in the first law is exactly the usual work$W=\int Fdx\rightarrow\int PdV$. For point particles, this is enough to completely specify the behavior of the system using Newton's first law, or energy methods. However, for macroscopic objects, the motion of the internal components (in thermodynamics these would be particles) have some additional degrees ... 2 The$W$term in the first law expression exclusively refers to the mechanical work done by a system and all other things , all other possible exchanges of energy are clubbed together in$Q$. Suppose I am the system under consideration , and I apply a force on a block and that does some mechanical work (that is the point of application moves a distance) ... 2 Kris Van Bael's answer has the right idea, and that is a problem for computer simulations. If you want to simulate this on a computer, you have to discretize time, and that causes problems with conservation laws. Another way to look at it is to look at an infinitesimally small amount of time (this is where Kris Van Bael's answer is going). The instant that ... -1 Many ways to prove why work done is zero: 1)I am pretty sure that your initial premise, namely "magnitude of the vector sum of the 'vertical' and 'horizontal' velocities is increasing" is flawed. You have written "no one increased velocity in vertical direction" is what is causing confusion (acc to me). I am sure that the vertical velocity is decreasing by ... 0 Kris's answer is correct. Here's another way to look at it. An object is moving at speed v. You exert a sharp force from behind. It's velocity increases. An object is moving at speed v. You exert a sharp force from the front (i.e. against the motion). It's velocity decreases. An object is moving at speed v. You exert a sharp force EXACTLY AT RIGHT ANGLES ... 8 This does not contradict newton, because the error is in the calculation: You are calculating in discrete time steps. In that case the calculated speed will increase, as if energy is not preserved. However, Newton's laws apply to continuous time. In the mathematical world in which newton laws are described, speed and acceleration are NOT defined by ... 0 A little trick required here. Perform a substitution first: $$mv \gamma(v) = u ,$$ hence your integral becomes $$\int \frac{du}{dt}dx ,$$ but notice that $$\frac{du}{dt} = \frac{du}{dx}\frac{dx}{dt}dx ,$$ but$\frac{dx}{dt} = v$- the definition of velocity! So the integral ... 0 I think I've understood it now .$ds=dr$. but$dr<0$and$|dr|=-dr$Because dr is a small position vector and position vector is directed along field . Now why I can't use ds directly is because the limits in the integral , (the upper and lower limit in integral notation) are in terms of position vector and not the displacement . Had they been in ... 2 New version The problem in your demonstration is when you write down$\vec{A}\cdot\vec{B} = ||\vec{A}||\,||\vec{B}||\,\cos\theta$. More exactly, in your case$||d\vec{r}||\neq dr$because$dr<0$when you go from$\infty$to$r$and a norm is positive by definition. So the sign error is introduced from 3rd to 4th line. Old version The demonstration on ... 1 Just to be clear, the potential energy of a particle of charge$q_2$at a distance$r$from a source of potential (supposidely at zero) of charge$q_1$is the work that an external operator has to provide to bring the particle from infinity to$r$at constant velocity. This reads then:$\int_{\infty}^r \vec{F}_{op}\cdot \vec{ds}$As people have said, the ... 2 When you calculate work, you do so along a given path. Here, that path has tangent vector$d\mathbf s$. This is a vector with direction; the minus sign will ultimately come from choosing the path's orientation--inward or outward. Edit: Aha, I think I've found the unintuitive part. The key is in the use of the coordinate$r$to parameterize the path, in ... -1 $$\mbox{d}\vec s = \mbox{d}r$$ Therefore, $$\vec F\cdot \mbox{d}\vec s= F\mbox{ d}r\mbox{ }\cos\theta=F\mbox{ d}r\mbox{ }\cos\pi=-F\mbox{ d}r$$ Edit: sorry for the error where I forgot to put the magnitude sign. I did mean the magnitude sign. $$\left|\left|\mbox{d}\vec s\right|\right| = \mbox{d}r$$ 0 There are two approaches to the definition of$C_V$, one of which displays its physical interpretation---$C_V=T \frac{\partial S}{\partial T}|_V$, i.e., as the partial derivative of entropy with respect to temperature under constant volume, scaled up by the temperature. A second one is less intuitive, but much easier to calculate and avoids some of the ... 1$\mathbf{r}$is a position vector and$\mathbf{s}$is a displacement vector between two points, let say A and B. In general case, they are not equal, but they can be if we properly choose the origin of the coordinate system: A={0,0,0} or B={0,0,0} The sign depends on at which point A or B the origin is placed. 2 Great question; I myself got confused for a moment there. I'm gonna try to be somewhat thorough, so bear with me. First consider the differential form of the first law of thermodynamics which holds for any quasi-static process. $$dE = \delta Q - \delta W$$ For an adiabatic process,$\delta Q = 0$by definition, so one obtains $$dE = -\delta W$$ On ... 0 Your intuition probably tells you that you waste more energy or better: work (e.g. your muscles), if you lift it up in a weird wavy way instead of lift it up in a straight line. This is certainly true! However, we don't talk about the things that make the object move but we talk about the object and the potential energy ($E_{pot} = mgh\$) of the object you ...

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