# Tag Info

## New answers tagged work

0

There are a lot of very physics-y answers here, so I'll give a more real-world example to help you wrap your head around the idea. Imagine (or better yet, try!) you have a weight on the end of a string and you are twirling it around your fingers. Once the weight is rotating, you let it wrap around your finger. You can see that as the object comes closer to ...

0

The centripetal force is directed radially inwards . Work done due to the centripetal force is 0 (as, S(displacement in one rotation)=0). Work done = F.S.cos(\theta)=m . a . cos(\theta).The accln. is thus directed tangentially outwards. Thus, tangential velocity increases.

0

When $\theta = 0$ the spring extension is zero and take this to be the zero of gravitational potential energy. With the rod at an angle $\theta$ from the geometry of the system find the extension $x$ of the spring and the vertical height $h$ through which the centre of mass of the rod has fallen. The potential energy of the system $V = \frac 12 k x^2 - ... 0 My guess:$\theta = 0$might not be equilibrium cause if you stop holding the A box, it will fall. at$\theta = 90$might be neutral equilibium (I assume the spring can pass through the wheel to the side of box A), since there is no tension in the spring, and the normal force on the B box balances the forces of gravity(assuming the B box is restricted to ... 0 The friction force doesn’t do work because the contact point of the disc with the ground changes. It is not same for a distance. At each moment, there is a new contact point. Hence, for all friction forces (for all contact points),$d=0$and$W=Fd=0$. 4 Notice that in the finite approximation the vector PR is not perpendicular to the radius, so work is done on it. By the drawing, this work is of the opposite sign that that in QR, so they both compensate to zero. I'll leave to you to compute both if you are really interested. Another way to see it, the gravity force is derived from a conservative field, so ... 1 Work done by a central force is Zero. At every moment the force is perpendicular to the displacement of the test particle. If you see your diagram it's very easy to see that at the final and initial position of the particle the force is not in the same direction. What he actually does is to assume that P and Q are actually arbitrarily close. So now the ... 0 If before putting the mass$m$on the piston, in absence of external constraints, the piston is in equilibrium; since it is not mentioned in the problem description that gas has received some amount of heat; thus it is impossible that the piston moves upwards after putting the mass$m$on it due to gas pressure. So, this question violates physics laws and we ... 0 Work, by its definition, is by mechanical. Potential energy and kinetic energy can be converted from work. We also talk, for example, heat using energy, which seems has nothing to do with either potential energy or kinetic energy. But later, you will find in statistic mechanics, heat is related to the kinetic energy of particles. But it is too expensive to ... 0 This may not be a definition of work. You need to put on more context. If the author was talking about how to get the work calculated, it is fair to make this statement. 4 The definition of work: $$W=\int \vec F \cdot\mathrm{d}\vec x$$ So, work requires a force and a displacement. That is all. Think of it like this: If you push hard on a wall, you might use much effort to apply large force - but nothing moves and no work of use is done. If you push against a balloon, you can make it move very far. But you didn't really do ... 0 A more general definition is ( the integral ) of force times distance. A static force, eg a body resting on a surface , does no work. The same body dropped has work done on it by the force caused by Gravity, potential energy being gradually converted to kinetic energy for the duration of the fall. 0 Yes Energy can be defined without reference to work. In the context of more advanced physics like Lagrangian mechanics, There's Noether's theorem; It states that for every symmetry that's present in our laws of physics, there exists a conserved quantity(does not change with time) that is associated with this symmetry. For example, if the laws of physics does ... 0 Joule, in his famous experiments, demonstrated the one-to-one relationship between mechanical work and internal energy change. Later, others demonstrated the equivalence between these and electrical work (and other forms of energy). 0 In kinematics one describes and analyzes trajectories, but not their connection to other physical processes. There is no force in kinematics. The concept of work uses the concept of force, so it does not belong to kinematics. It belongs to dynamics, a study of origins of changes in the motion. if it is then why is it used to define energy? can energy be ... 0 Force The tendency to make something accelerate. Newton's 2nd law: $$\sum \vec F=m\vec a$$ Physical strength Can be different things. E.g. hardness$H$: a materials resistance against "bumbs" and indentations in the surface, yield stress$\sigma_y$: the stress a material can withstand before starting to deform ultimate tensile stress ... 0 Force, energy, and work are all just different ways for us to describe motion. Energy (more specifically kinetic energy) is a measure of how fast an object's motion is. If an object has non-zero velocity, it has energy by definition. Force is a measure of how fast an object's rate of motion is changing. If an object's velocity changes, it's experiencing a ... 1 We have to talk about Physical Strength last, for two reasons: (1) we have to clearly define the other physics terms first, and (2) Physical Strength gets us into biophysics so we'll have to talk about how muscles generate force, etc. "Force" is a push or a pull or a twist (if a twist, then it is also called "torque"). Force is what is required to ... 2 Your original equation is incorrect. For a process occurring in a closed system, the equation should read$$\Delta U=Q+W$$where U is the internal energy of the system, Q is the heat added to the system, and W is the work done by the surroundings on the system,$W=-\int{PdV}$. If the process takes place at a constant pressure, then$W=-P\Delta V$, and the ... 1 The symbol$\Delta$(here) simply means difference, e.g.: $$\Delta H=H_2-H_1$$ So, enthalpy of the end state minus enthalpy of the initial state. In the case of an exothermic reaction the system has lost enthalpy, so: $$H_2<H_1$$ Thus, for an exothermic reaction: $$\boxed{\Delta H=H_2-H_1<0}$$ It is possible then for$ΔH$to be negative (if ... 0 I guess the puzzle is to transfer heat out to surrounding and, at the same time, do work to the surrounding. This doesn't violate the first law for sure as the energy is conserved by decreasing internal energy. And this doesn't violate the second law of thermodynamics as well. 0 Consider the water liquid at its freezing point. When water freezes its internal energy goes down - bonds are made. The decrease in internal energy is equal to the heat removed from the water and the work done by the water in expanding against its surroundings. 0 There is no inconsistency. The first law of thermodynamics tells you that during any process the internal energy$U$of any simple compressible system will satisfy$\Delta U = W + Q$where here$W$is the heat received by the system and$Q$the heat received by it. Upon changing from water to ice, the part of the system that undergoes the phase change does ... 0 It does work because it operates like a Heat Engine where the energy travels from something with higher energy (water) to something with less energy (air) to create the work done by the ice. The air has to be colder than the water and below 0C in normal conditions for the water to even freeze. It is just another Heat Engine-like working. 0 We are talking about maximum amount of work, so you still consider an ideal (Carnot) cycle. But the efficiency is changing as the tank cools down, so there is an absolute maximum amount of work that can be extracted from this heat engine. Efficiency of Carnot engine is$\eta=1-\frac{T_{min}}{T_{max}}$, and is defined as work over heat transferred at the hot ... 1 Force is a vector. Potential energy is a scaler. Forces which have associated potential energy functions as called conservative forces. Conservative forces act in such a direction that, if released from rest, the potential energy function associated with that force will decrease (and the kinetic energy will thus increase) with the velocity increasing, until ... 0 What is the significance −ve sign here? And will the work done be +ve? If I want to lift some mass upwards I need to apply a force at least equal to it's weight. This means if I want to pull the mass upwards, with a certain acceleration then an additional force has to be supplied along with it's weight. But the resultant force will be the difference in ... 2 From the definition of work $$W = \int dx F$$ and $$P = \frac{dW}{dt}$$ you can see how we can arrive at $$P = F \frac{dx}{dt} = F v$$ (when considering only the absolute value). To understand it intuitively, imagine the case of a frictionless system in which the car can move at a certain speed without any opposing force. The power required to keep it at ... 4 Walking requires raising and lowering the centre of gravity, as well as moving the limbs. Both of these require muscles to contract and extend and use energy. If the muscles were perfect springs then the energy stored during contraction could be fully recovered during extension. No energy would be lost. But they are not perfect springs - some energy is ... 1 To understand why holding objects costs energy even though the work appears to be zero, you have to understand how muscles work. When you are holding an object, your muscles are contracted. The process of muscle contraction consists in a protein filament called Myosin pulling another filament, called Actin. Since this is a dynamical process (the Actin ... 0 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 ... 0 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 ...

Top 50 recent answers are included