New answers tagged

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Kinetic energy is $$K = \frac{1}{2} \vec{\omega} \cdot [I] \vec{\omega} $$ when the 3×3 mass moment of inertia matrix $[I]$ is expressed in world coordinates. Remember $$[I] = [R] [I_{body}] [R]^\top$$ is how body inertias is transformed into world inertias. You seem to apply a scalar mass moment of inertia to a vector rotation. If you are careful with the ...


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Suppose that you double the separation of the plates of a parallel plate capacitor of initial capacitance $C$ which is connected to a battery of emf $V$. The capacitance of the capacitor become $\frac C 2$ and the energy stored in the capacitor changes from $\frac 12 CV^2$ to $\frac 1 2 \frac C 2 V^2$. There is a decrease in the energy stored in the ...


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Assuming no air friction, you compute the initial velocity needed from conservation of energy: $$ \frac12 mv^2=mgh $$ The impulse needed is $mv=F\Delta t$. The product of these ($F, \Delta t$) is constant - shorter time implies higher force. The above assumes the time of impact is short enough not to affect the over all time (otherwise you need to solve ...


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The potential energy is the ability to perform work in the way that... potential energy $U$ is converted into kinetic $K$ energy during the fall. The kinetic energy implies a speed $v$ and thus causes a momentum $p=mv$. On impact this momentum is reduced drastically causing a force $F=dp/dt$. Is the impact-surface displaced by x, then the work $W=Fx$ has ...


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To my understanding, work is done on object A when object B is applying a force on object A, causing object A to displace. Work is done whenever a force displaces an object. Since energy is the ability to do work, what work does a moving object do, due to its kinetic energy? A moving object might not do any work at all. Imagine an empty ...


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The work the ball that you kick does is the ball going against air-resistance and changing its velocity.


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Potential energy is associated with a system, not a particle. One must calculate potential energy for pairs of interacting particles, not individual particles. $F_{ext}$ is a force external to the system. It can do external work and raise the energy of the system. $F_R$ is work internal to the system, and does not increase the energy of the system. But ...


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It's true that there are many inner products you can choose on $\mathbb{R}^3$. However, physics supplies the additional principle of rotational invariance: the result should not depend on our coordinate system. Now, any inner product of vectors $a$ and $b$ can be written as $$a \cdot b = a^T M b$$ for a matrix $M$. Rotational invariance tells us that $M$ ...


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There is a little bit more thinking behind saying that $P=\vec F \cdot \vec v$ than it being a generalised multiplication in 3D. There are even cases where multiplication with scalar becomes a cross product when using 3D vectors. For example, torque $T=Fr$, becomes $\vec T = \vec r \times \vec F$. Whenever implementing vectors into existing scalar equations, ...


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The equation $\Delta U = Q - W$ is complete in itself. The confusion arises in the definition of enthalpy. We tend to think that the "pressure energy" $PV$ and the work done by the system $W$ are somehow different. But the fact is that there isn't a well defined physical meaning to the $PV$ term. It is confusing because when we add $U$ and $PV$, we think ...


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A non conservative force is generally anything that is affected by friction or air resistance. The amount of energy lost is distance dependent because the further the path traveled, the more an object is affected by these forces. Conservative forces are displacement dependent, meaning they depend only on an initial and final position, and that all the ...


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The simplest example for a non-conservative force is frictional force. For example, to move an object from point A to point B on the table we would require say $10$ J of energy, now to put it back to point A, we would need same $10$ J of energy. Practically nothing observable has changed, but we've performed $20$ J of work, which went to overcome frictional ...


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Since work done by a force $\vec F$ undergoing a displacement $d\vec r$ is defined as $\vec F \cdot d\vec r$ when this dot product is positive the force and displacement are in the same direction and is negative when they are in opposite directions. The work done by a frictional force does not always have to be negative. Imagine a block $A$ on top of block ...


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Force is a vector, meaning magnitude and direction. Work done by a force is relative to the direction of a force is the scalar value obtained by performing the vector dot product of the force and the displacement (which is also a vector). If something isn't coming out to what you expect when you compute work, make sure you have the right magnitude and ...


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Are isothermal free expansion and adiabatic free expansion different? No. They are the same. Your mistake is in thinking that $PV^\gamma = \text{constant}$ applies to a free expansion. That expression is for a reversible (i.e., isentropic) adiabatic process. A gas that has undergone a free expansion has more entropy after the expansion is complete than ...


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You can not classify free expansions into any of those categories as free expansion is not a reversible process and hence the intermediate states are not well defined. The equations are not working because they find the area under the p-v graph but here no such graph can be made.


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If you study the problem in the frame of reference of the CM, the contact force of the plane or the rope does work, because the point of contact is moving in this frame. However, note this frame is not galilean, so you may not blindly apply the work-energy theorem (in this exact case it is valid, but take the good habit not to apply it in non-galilean ...


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As an answer to your second question. If you released the dielectric it would reach the equilibrium position and then overshoot as it would have kinetic energy eventually stopping on the other side and then returning. Compare this with the oscillation of a spring-mass system. So the dielectric would undergo oscillatory motion and if damped it would ...


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A force field is called conservative if its work between any points $A$ and $B$ does not depend on the path. This implies that the work over any closed path (circulation) is zero. This also implies that the force cannot depend explicitly on time. Consider for instance a time decaying force on a straight line. Choose a long closed path. The magnitude of the ...


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According to Newton's third law, the javelin exerts as much force on the athlete as the athlete exerts on the javelin. During the throwing phase (when the athlete is in physical contact with the javelin), the athlete exerts a force on the javelin. That force acts only on the javelin. At the same time, the javelin exerts an equal and opposite force on ...


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The javelin does work on the athlete's arm/hand equal in magnitude and opposite in sign to the work that the athlete's arm does on the javelin. Both the arm and the javelin experience a force applied through a displacement.


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I'm considering this a "Athlete throws javelin" situation. As you correctly put it, the javelin exerts a force on the athlete. However, by the very definition of "work", you can only say the javelin does work on the athlete if the point of application of the javelin's force moves in the force's direction. If the point of application wasn't displaced, then ...


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You should plug in the horizontal force acting on the object if you want to know the work done by this force. Due to friction heating up the plane not all of this work is converted in the objects kinetic energy. PS: I assume you meant "while calculating the work" rather than "while calculating the force"..


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Work done is $\vec F \cdot \Delta \vec x$. If $\vec F$ and $\Delta \vec x$ are in the same direction then the work done is positive. If $\vec F$ and $\Delta \vec x$ are in opposite directions then the work done is negative. Consider a spring fixed at one end as a system and an external force $\vec F$ stretching the spring. If the external force $\vec ...


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Potential energy of a body is its capacity to do work by virtue of its position in a conservative force field.So if the body is free to move it will do so in such a way as to reduce its potential energy and the reduction in the potential energy will be equal to the positive work done by the body which could result in the raising of a weight or displacement ...


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It is just a convention. If we would use the opposite convention, we would get for conservative forces $$\vec F=+\vec\nabla U.$$ You can easily see (think in the one dimensional case) that the particle would move to points of maximal potential energy. This only sounds strange because we are used to the opposite. The mechanical energy principle would retain ...


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The decrease or increase in potential energy is converted as mechanical work either to increase or decrease it's kinetic energy (we are considering here conservative systems). Suppose you have a system at rest at some height above the ground level, say, a ball placed on the top of a mountain. The work done in order to bring the ball of mass m to a height h ...


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It's a convention. The real reason is so that we can have: $$\begin{align} \Delta {\rm KE} &= W_{\rm ext}\\ \Delta {\rm KE} &= W_{\rm con} + W_{\rm noncon}\\ \Delta{\rm KE} &= -\Delta {\rm PE} + W_{\rm noncon}\\ \Delta{\rm KE} + \Delta {\rm PE} &= W_{\rm noncon}\\ \Delta {\rm E} &= W_{\rm noncon} \end{align}$$ Which doesn't work ...


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It would be better to say that potential energy is the amount of work that a system can do. Say you have a system consisting of two masses - a brick and the Earth. As the brick moves down U decreases, the force pulling the brick and the Earth together acts downwards on the brick and it can do some positive work. On the other hand, if brick moves up, U ...


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Work is calculated by force multiply by distance. So the work that you want to calculate is: That d is the distance the box is pulled.


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Potential energy is not energy. If you, for example, place an object of mass m on your outstretched palm and raise it through a distance h, so that it is initially stationary and ends up being stationary, then the total work done on the object by the weight force acting on it is -mgh and the total work done on it by the normal force acting on it due to its ...


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An electric field is the force that fills the space around every electric charge or group of charges. Electric fields are caused by electrical forces. Electrical forces are similar to gravitational forces in that they act between things that are not in contact with each other. Electric fields are also analogous to magnetic fields resulting from forces acting ...


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By work-KE theorem, $$\Delta \text{KE} = \text{Work done by NET force}$$ $$=\text{Work done by} F_1+\text{Work done by} F_2+\cdots$$ Now if a certain force $F_i$ is conservative, you have the choice of defining its corresponding PE so that $$\text{Work done by}F_i=-\Delta \text{PE}_i$$ and MOVE it to the left hand side so that you have ...


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The answer is to do with what happens when you stop moving the object upwards. You say that the upwards velocity is 'low enough that it does not leave your hand': the only value of velocity for which that is true is $0$. What this means is that, if at time $t_0$ the object is being moved upwards with velocity $v_0$ and is at $h_0$, and you suddenly stop ...


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You are not defining your systems clearly enough. If the system is the mass and the Earth then in the situation that you have described the external force doing work on that system increases the potential energy of the mass-Earth system. Now look at the system which is the mass alone. Then there are two forces acting on the mass. The force as described ...


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If you lift a book, then you are doing work on it. You are actually spending work on raising it's kinetic energy and on lifting it up. $$U=K+W$$ When you reach some height and stop, then all the work done is what determines your height. If you move your hand faster in order to speed up the book to a larger kinetic energy, then you've spent more work on ...


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John Rennie has explained the problems with units here. Now, you'll burn about a factor 4 more than the work you perform, due to losses when glucose or fats are burned to allow the muscles to do the work. The Gibbs free energy change when glucose or fat reacts with oxygen and changes into water and carbon dioxide gives you the maximum amount of work that can ...


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Remember that every time we talk about work being done, we must know which force that is doing it as well as how the signs are defined. where $W(begin)$ is the Work done to bring the object from infinity to a certain position in space, and $W(after)$ the contrary, from position to infinity." What force is doing this work? That would be some external ...


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There are two equivalent ways of defining the potential at a point. The potential at a point is the work done by an external force in bring unit positive charge from the zero of potential (often taken as infinity) to the point. The potential at a point is minus the work done by the electric field in bring unit positive charge from the zero of potential ...


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Due to an accident of history there are two different units called the calorie and the Calorie - yes, the only difference is the capital C. The calorie is 4.2J but the Calorie is 4.2kJ, and the calories counted in diets are actually Calories even though they are invariably written on the food packaging with a small c. So you only used 3 Calories walking up ...


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The solution to this problem needs to take into account the fact that the ramp is frictionless, and the mass is moving at constant speed. None of the solutions presented so far have taken this into account. If $\theta$ is the angle that the ramp makes with the horizontal, the component of F along the ramp is $F(\sin53\cos\theta+\cos53\sin\theta)$. If the ...


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As you know, work is calculated by force product distance that parallel to force. So the work that you want calculates below:


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Work (or energy) is transferred from one particle to another, but the net effect is no change overall. How? Consider a collision force acting between two particles over a small time frame. During that time frame the particles move, and the work done on one particle is ${\rm d}W_1 = \boldsymbol{F} \cdot {\rm d}\boldsymbol{x}_1$. Since an equal and opposite ...


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The collision doesn't happen at a single point in space - rather the colliding objects exert a forces on each other over a distance as they approach and the recede. Consider a tennis ball hitting a racket - the ball and the strings of the racket deform and we get an increasing elastic restoring forces until the two objects at at their closest approach. ...


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The reason is that work always depends on the actual velocity the body has during its movement, even if you compute the work for only one force out of many. In other words, $W_{F_i} = \int F_i\cdot v dt$, rather than, as you might perhaps expect, $W_{F_i} = \int F_i \cdot v_i dt$, where $v_i$ is a hypothetical "what $v$ would be if only $F_i$ acted". Now ...


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It's not that it's not possible, it's just not necessary. What we seek is not the sum of the changes in kinetic energy. We seek the total change in kinetic energy. Why would the former be of use? (in Newtonian mechanics that is, since my knowledge is limited to it as of now) If a box is accelerated through the force of $F$ through a distance $d$, from rest, ...


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$\Delta K$ means $K_\text{end}-K_\text{start}$. How would you split this in parts?


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1) What can be say about the work done by the man to the weight? Unlike gravity, the force exerted by the man is in general not constant, does not depend only on position of the body (you may apply different force on the weight at the same height $h$ on the way up vs. on the way down), and is not conservative (does not arise from a potential). When the ...


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1&2) Let's suppose there are only two forces in play: the one exerted by the man, and weight. The sum of works done on the body equals the variation of kinetic energy (this is a theorem of mechanics; I fail to find its English name). Since the body has no velocity at the start and at the end, the sum of works is zero. The net work done by the man on ...


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The correct statement is - the total work done on the weight is zero i.e. the total energy of the weight before and after the experiment is same. However, when the man is lifting the weight he is obviously working against gravity. More importantly, when he is lowering the weight, he is still working against gravity, as gravity would rather lower the weight ...



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