I have learnt that it states "The work done on a body is equal to the change in kinetic energy of the body" which we derived by considering a position-dependent force. However, if mechanical energy is not conserved and one can do work by friction forces, one can increase the internal energy of the body. From my understanding the work-energy theorem is thus not valid unless we consider the internal energy as kinetic energy.  Is the work-energy theorem only valid for conservative, i.e. position-dependent forces?
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How can the  be valid in presence of non-conservative forces since conservation of energy is not there?
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This may be a very basic question but I am not seeing how it works. Consider the standard example of an ice skate rotating about his/her center of mass and pulling in his/her arms. The torque is zero so we have conservation of angular momentum. This implies that $\omega$ increases to keep $I\omega$ constant, but then $K_{rot}=\frac{1}{2}I\omega^2$ doesn't stay constant, it increases. This implies that there is work done, but what force is doing this work?
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So I have some confusion here, I am sure I knew this at some point.  Let's say an object of 10 kg is dropped from a height of 10 m.  When it reaches the ground, the work done on the object should be the force ($mg$) x distance or 10 kg x 9.8 m/s/s x 10 m.  That gives 980 joules of work done on the object by gravity.  But the object did not gain 980 joules of .  It lost 980 joules of GPE and gained 980 joules of kinetic energy (up to the point of it reaching ground level).   Using the change in GPE and KE, it looks like no work was done on the object because the loss in GPE equals the gain in KE.  So   a) am I right that no net work was done on the car by sum of all forces?    b) is the work done by gravity equal to force time distance, or is it equal to the change in mechanical energy of the object which is zero?
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I'm in a mechanical physics class, and I'm having a hard time understanding what the quantity of work represents. How can I understand it conceptually?
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Consider the dynamics of a particle P shown:  Newton's second law states that: $$\frac{d}{dt}(m\dot r) = \mathbf F$$  where, $\boldsymbol{r}$ is the position vector of the particle, $\boldsymbol{\dot r}$  is its velocity vector and $m$ is its mass. $\mathbf F$ is the total external force (vector) acting on P. Prove the following result regarding the existent of mechanical energy integral of motion of the system:  Theorem: For the dynamical system, if $F = -\nabla V (\boldsymbol{r}) $ i.e the force, $\mathbf F$, is the negative gradient of a scalar function $V(\mathbf r)$ of only the position vector, then the total mechanical energy of the system is conserved. Moreover, the function $V(\mathbf r)$ is nothing but the potential energy of the system.  Hints:  The above theorem states that if  $F = -\nabla V (\boldsymbol{r}) $ then the mechanical energy is an integral of motion.  The gradient of a scalar function, $V$ is written as:  $\nabla V = \partial V / \partial \boldsymbol{r} $  Also, the gradient of a scalar quantity is a vector. Do not worry about writing it in terms of the components  Recall the mechanical energy is the sum of KE and PE. Mechanical energy is given by the standard form:  $$T=1/2 ∗ m\boldsymbol{\dot r} \cdot \boldsymbol{\dot r} $$  where $\boldsymbol{\dot r} \cdot \boldsymbol{\dot r}$ represents the dot product of $\boldsymbol{\dot r} $, which is the velocity vector, with itself (i.e. square of the speed: $\boldsymbol{\dot r} \cdot \boldsymbol{\dot r} = v^2$ )  Also, let $G = G(\mathbf r, t)$  be a scalar function of a vector, $r$, and a scalar, $t$. Then, the total derivative of $G$ is related to its partial derivatives by:  $$dG = \frac{\partial G}{ \boldsymbol{\partial r}}\cdot d\mathbf r + \frac{\partial G}{\partial t}\cdot dt$$  with $\frac{\partial G}{ \boldsymbol{\partial r}} $  identified as the gradient of G, which is a vector quantity, and $\frac{\partial G}{ \boldsymbol{\partial r}}\cdot d\mathbf r $  as the dot product between the gradient of $G$ and $d\boldsymbol{r}$  Do not assume that $V(\boldsymbol{r})$ is the potential energy as this should be a natural conclusion of the analysis.  To begin, take the time derivative of the kinetic energy. Recall that if $c = \boldsymbol{a} \cdot \boldsymbol{b} $ is the dot product of two vectors, then the time derivative of c is: $$\dot c = \boldsymbol{\dot a} \cdot \boldsymbol{b} + \boldsymbol{a} \cdot \boldsymbol{\dot b}$$  ASSUME THE PARTICLE'S MASS, $m$ is constant  Edit: Attempted solution: $$\frac {d}{dt}(m \dot r) = F $$ $$\frac {d}{dr}(m \dot r)\frac {dr}{dt} = F$$ $$d(mv)v = Fdr$$ $$\int mvdv = \int Fdr$$ $$\frac {1}{2}mv^2 + \int-Fdr = C $$  Seems to prove the above theorem, however this kind of seems to ignore the fact that the variables are vectors. How can a similar approach be applied to the vectors? And still derive the potential energy at the end?  Basically, I'm asking how the potential energy can be an outcome of $$\frac {d}{dt}(\frac{1}{2} m \dot r \cdot \dot r) $$ and $$\frac{d}{dt}(m\dot r) = \mathbf F$$
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Work can be calculated by $W = Fd\cos\theta$ where $\theta$ is the angle between the force $F$ and the displacement $d$  Let's say there's a ball and a (physically ideal - no friction etc.) robotic arm situated in otherwise empty space. The arm takes the ball, moves it around in a circle, and then returns it exactly to where it started. The arm also returns to its starting position.  In this case, there is no displacement overall. The ball and the robot arm are in the exact same positions as when they began.  Thus, $W = F\cdot 0 \cdot \cos\theta = 0$ So no energy was required to move the ball in a circle.  However, this disagrees with my intuition, because if I were to make a robot arm that would do this, I feel like I would need to give it an energy source (for example, a battery), and that by the time the robot was done, I would have lost energy from that battery.  Does the movement require energy? If so, where did I mess up in my calculation? If not, why would my robot need a battery pack or some energy source in order to make this movement? (If it doesn't, please explain.)
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My understanding about  is a force that its work is independent of path such that we can construct another form of the work called potential to make our life easier.  For friction, if I start from microscopic point of view, it should be the macroscopic effect of the electric force or gravity which are both conservative force.  Why do we initially have description by conservative forces (electric/gravity force) but end up with a macroscopic description of nonconservative force(friction)?
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What exactly is ? My book confuses me:     a force can lift an object to a height h, or it can accelerate an object through gravity. In all these cases, a force displaces an object and change the object's total energy.   The examples it gives confuse me. On one hand, there is lifting and on, accelerating through gravity.   I can imagine how lifting will change total energy (kinetic + potential). It essentially gives it more potential energy, and does not take from it any kinetic energy. But just dropping an object and let gravity do the work does not seem to me to increase total energy. Because the total amounts of kinetic and potential energy should be equal.  If just let drop, the object will gain kinetic and lose some potential. Not really anything added to the total energy.
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According to 2nd law of thermodynamics, it is not possible to convert heat 100% into work.   Well, investigating about the Carnot Cycle, I have found out that in the first step, the process is to heat a gas without rising its temperature, this is, converting all the heat into work. Isn't this a violation of the 2nd law?  I clearly don't understand something about the law or about the Carnot Cycle. I would like to know where I'm failing in my understandig. Thank you.
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Kinetic energy conservation in collision.  In any type of collision type of collision if we take the system to comprised of both bodies then the net external force is zero. So the work done by external force is also zero. By the relation $W=K_f-K_i$ as $W$ is zero hence kinetic energy is conserved. But this not the case.   Can you tell me what is the flaw in my argument? Some people say that the argument has neglected the fact that some kinetic energy is transformed into heat. But heat is created due to the interactions between bodies during collision which is thus an internal force and thus shouldn't affect kinetic energy.
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How is it proved to be always true? It's a fundamental principle in Physics based on all of our currents observations of multiple systems in the universe. Is it always true to all systems? Because we haven't tested or observed them all. Would it be possible to discover/create a system that could lead to a different result? How are we 100% sure that energy is always conserved? Finally, why did we conclude it's always conserved? What if a system keeps doing work over and over and over with time?
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I have already read        and many others   But I still have the following problem:   On one hand I've been told that if pure rotation takes place then the object goes on forever. This must imply no frictional force in any form acting on the body. On the second hand I've been told that during pure rotation(no external forces like wind) on a plane surface friction acts opposite to velocity of center of mass (accompanied by change of normal) and slows it down.     Both seem logical enough: The point of contact doesn't slip therefore no friction is required; So no retarding force and it must go on. Things don't go on forever; Friction must be present to slow it down.    So, which hand is correct? Or is there a third hand altogether?
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If the displacement is in the direction of the force, then work is positive. If the displacement is in the opposite direction of the force, then work is negative. If the displacement is perpendicular to the direction of the force then work is zero.  So, the work done depends on direction. So work should be a vector quantity. Then how is work a scalar quantity?
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Almost every book on physics that I read have some weird and non-clear explanations regarding the potential energy. Ok, I do understand that if we integrate a force over some path, we'll get a difference in some origin-function values ($\int_{A}^{B} Fdx = U(B) - U(A)$). This function is the potential energy. Of course, whether we can define this term or not depends on the force.  Now, here's an example of explanation (to be more precise - lack of explanation) regarding the GPE from one of the books:     "...When a body moves from some point A to point B, gravity is doing work: $U_A-U_B=W_{A \to B}$. The magnitude can be calculated using an integral:   $W_{A \to B}=\int_{r_A}^{r_B} F(r) dr = \int_{r_A}^{r_B} \left(-\frac{GMm}{r^2} \right)dr=(-\frac{GMm}{r_A})-(-\frac{GMm}{r_B})=U_A-U_B$      ...      Thus,  when $r_A&gt;r_B$, the magnitude is positive and therefore $U_A&gt;U_B$. In other words: when the distance between the bodies is being increased - the gravitational potential energy of the system is also being increased.   Giving absolutely no explanation on why all of a sudden they put a minus sign into the integral.  From another book:     Work done by Coulomb's force: $W_{el}=\int_{r_1}^{r_2}\frac{q_1q_2dr}{4 \pi \epsilon_0 r^2}=\frac{q_1q_2}{4 \pi \epsilon_0 r_1}-\frac{q_1q_2}{4 \pi \epsilon_0 r_2}$      ... Calculating the work gravity is doing is no different from the calculation of the work done by an electric field, with two exceptions - instead of $q_1q_2/4 \pi \epsilon_0$ we should plug $G M m$, and we also should change the sign, because the gravitational force is always a force of attraction.   Now, this is not satisfying at all. So what if it is an attraction force? How this should influence our calculations, if the work is defined as $|F| |\Delta x| \cos \theta$, so the sign only depends on the angle between the path vector and the force vector? Why they put a minus sign? Is is just a convention or a must thing to do?  Some say the sign is important, others say the opposite. Some explain this as a consequence of that we bring the body from infinity to some point, while others say it is a consequence of an attractive nature of the gravitational force. All of that is really confusing me.  Also, in some of the questions like "what work is required to bring something from point $A$ to point $B$ in the field of gravitational/electric force", the books sometimes confuse $U_A-U_B$ and $U_B-U_A$ - as I understand it - the work that I must do is always $U_B-U_A$. However the work that the force that is being created by the field do is always $U_A-U_B$, am I correct?
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Suppose a block of mass $m$ is being pulled on a hill by a force $F_{app}$, the block is being pulled slowly such that $\Delta KE = 0$.  Our teacher showed that the expression of work done by friction $W_{fric}$ is independent of the path traversed by the block, which is not a characteristic of non conservative force. How is this possible?  Working: $$ \Delta KE = 0 $$ so,$$W_n + W_g + W_{fric}+ W_{app}= \Delta KE = 0 $$(from work-energy theorem).  As $$W_n=0$$(work done by normal Force), $$0+W_g +W_{fric} + W_{app} =0\, .$$ Now, $W_g$ is independent of path and is equal to $-mgh$ so, $$W_{app} = -(mgh + W_{fric})\, .$$ Now for $W_{fric}$, $N=mg \cos(\theta)$ where $\theta$ is the inclination of hill slope wrt positive $x$-axis, $$  dW_{fric} = kmg\cos(\theta) cos(180) ds\, , $$ (from $W = fs\cos(\theta)$ and $F_{fric} = kN$.) $ds$ is small displacement along the slope, so $ ds \cos(\theta) $ is small displacement along positive $x$-axis, Reordering our last equation, $$ dW_{fric} = -kmg dx$$ as $\cos(180) = -1$ and $ds \cos(\theta) = dx$ Integrating on both sides, $$\int{dW_{fric}} = -kmg \int dx$$ so $W_{fric} = -kmgx$ which does not depend on the length of path taken but only on the horizontal displacement $x$.
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As I am seeing this,  the man and boat will have different displacement from ground frame,  so there should be a net work done by friction .  The books says,  total work done by friction is zero when friction is static and internal to system. Please explain..
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I have a question regarding work done / energy done. Say a $70\mathrm{kg}$ man walked $100\mathrm{m}$ on a horizontal surface as a constant speed, is it correct to assume the energy done by this man would be: $$\text{Energy} = \text{force}\cdot\text{distance}=70\cdot9.81\cdot100=68,670\text{ joules}$$ Where the force is just the weight of the man. Then say he walks another $100\mathrm{m}$, but this time he ascends $10\mathrm{m}$ and descends $10\mathrm{m}$ during the walk. To calculate the energy would you then have to add in energy done against gravity, so: $$\text{Total energy}=(\text{force}\cdot\text{distance})+(m\cdot g\cdot h)=(70\cdot9.81\cdot100)+(70\cdot9.81\cdot10)=86,670+6,867=75,537\text{ joules}$$