![]() ![]() This issue was also addressed by Miller in his earlier response. The piston is treated as a rigid body, and a rigid body can have no change in its internal energy (e.g., a rigid body cannot be compressed.) The bottom of the piston is exposed to the gas and the top is exposed to the atmosphere. ![]() Let the system be defined as the gas in a container, a closed thermodynamic system of constant mass, with a moveable boundary, that being the interface of the gas with a moveable piston on top of the gas. The answers to both these questions follow, using a detailed evaluation of the work done on the piston by a single force, from the pressure of the gas, contrasted with the net work done by the total force on the piston, the vector sum of all forces on the piston: the force from the gas pressure, the weight of the piston, the external atmospheric pressure on the piston, and other forces on the piston due to external loads driven by the piston. And, if there is also an external pressure on the piston (e.g., from the atmosphere), and from other forces on the piston due to external loads driven by the piston, why are they also not included in the work done by the gas? The answer is: we are interested in the work done by a system on its surroundings for the piston example, this is the work done by the gas (the system) on the piston (the surroundings), but this is not the total work done on the piston. The angle between the objects motion and the force in such a case is 180 and. Find AB AB jAjj Bj cos 5:15 8:55 cos46:0 30:6 0.4 A force F (8i 3j)N acts on a particle that undergoes a displacement r (2i j)m Find the work done by the force on the particle. With energy leaving the object, the work done on the object should be negative. The two vectors make an angle of 46:0 with each other. Specifically, why is the weight of the piston not included in the work done by the gas? Based on comments and answers by others, there is confusion about the answer to this question. Vector Ahas a magnitude of 5.15 units, and vector Bhas a magnitude of 8.55 units. The detailed evaluation also answers a question by in an earlier comment for a gas expanding and moving a piston. For a quasi-equilibrium process this work can also be evaluated considering changes in the internal pressure and volume of the gas but in general for a non quasi-equilibrium, irreversible, process the work cannot be evaluated using changes in the internal pressure and volume in the gas, because the gas is not in a definite state.Ī detailed evaluation of work for a gas expanding and pushing a piston follows. The dot product above ensures that we calculate the product of the component of the force (in the direction of) parallel to the direction of the displacement.The answer to the question by the OP is that the work done by a system can always be evaluated considering movement against the external force (pressure). So what you end up doing is calculating the following integral $$W_$$ not very small displacement) will be the sum of the work done over those infinitesimally small displacements. The work over a short segment of path is the dot product between force $\vec F$ and small displacement $\delta r$. ![]() I'm fine to solve these problems, but I really want to understand the "why". I apologize in advance for the simple (simplistic) nature of these questions. So my two questions are 1) why use the tangent vector, as opposed to some other way to estimate arc length and 2) why the unit tangent vector? The unit of work in SI units is joule (J) which is defined as the amount of work done when a force of 1 Newton acts for distance. The amount of work (W) done on an object by a given force can be calculated using. But why do we take the dot product of the force and the unit tangent vector, as opposed just to the 'tangent vector'? vector whether the work adds or removes energy from the object. Why do I use the unit tangent vector? Is it because the distance being traveled is so infinitesimal that the tangent vector (a straight line, informally put) can approximate the curve? I suppose this makes the calculation easier? And why do we use the "unit" tangent vector? Again, I know how to find the unit tangent, and I understand what it represents generally. I can solve these problems, but I have a question about the intuition. I find it written many places that "you can find the work along a short segment of the path by taking the dot product of the force and the tangent vector." ![]()
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