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## New answers tagged work

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Kinetic energy tells us how much work is required to stop a body. The total work done on a body equals its change in kinetic energy. That's what the kinetic energy theorem says. The "work of a body" makes no sense. Work is something done ON a body BY an agent. It can be seen as an exchange of energy from that agent to the body. For instance, if you make a ...

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The interesting thing is that in the diagram shown above the frictional force is zero so there is no way a frictional force can do any work. If there was a frictional force acting on rolling object then there would be a torque about its centre of mass and so there would be an angular acceleration. Also if there was a frictional force acting on the rolling ...

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The point in which the friction acts is not actually moving, but rather as the object rolls past any given point, a new point of friction takes over. See in the animation below how a point on the circle touches briefly:

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Power = $\vec F \cdot \vec v = Fv$ if $\vec F$ and $\vec v$ are in the same direction. In this case the power at a given time is instantaneous force $\times$ instantaneous velocity. In the first case you are finding the average power which you will note is equal to half the final instantaneous power which is second case value.

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When you are walking you are doing work against gravity and friction. Consider this - when you walk on a flat surface, you shift your body weight on to say right leg. Lift the left leg and move it by a step. For the next step, you shift the weight on the left leg, lift the right leg and move forward. Thus you move. What is the work done? The leg consists ...

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Using the first $2$ statements or conditions, apply the principle of conservation of energy and you will be able to calculate the resistive force offered by the target material. Next use the second condition and again apply the principle of conservation of energy to get the required answers, mainly the velocity of the bullet after penetrating the target.

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If A would gain kinetic energy, it would move far from B. As A would move more far, Potential Energy of B won't increase as distance had increased proportionally.

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There is no conflict here. Let the two charged particles ($M,Q$) be the system with no external forces acting. Momentum is conserved and so for all time $M_BV = M_A V_{Af} + M_BV_{Bf}$ Energy is also conserved and so for all time $\frac 12 M_B v_B^2 + \dfrac{kQ^2}{R_i} = \frac 12 M_B v_{Bf}^2 + \frac 12 M_A v_{Af}^2 + \dfrac{kQ^2}{R_f}$ The ...

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Potential and potential energy are defined for pairs of objects, not individual objects. It's meaningless to say "the potential energy of A". One must say "the potential energy of the system consisting of A and B". There is only one potential and potential energy in your problem. Perhaps the confusion comes from the way potential is introduced in ...

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You know that the centripetal force is given by $\vec F_z = m\omega^2r \, \vec e_r$ ,where $\vec e_r = \cos \theta \, \vec e_x + \sin \theta \, \vec e_y$ is the unit vector in radial direction. We want to calculate the work given by the line integral $$\int_C \vec F_z \cdot \mathrm d \vec r$$ where the position of the point mass $\vec r = r\, \vec e_r$ ...

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I came up with a solution seeing John Rennie's comment. The centripetal force, $\vec F= -F \hat r$ so infinitesimal work done by centripetal force, $$dW=\vec F.d \vec r= -F \hat r.d\vec r$$ but, $\hat r⊥d \vec r$ so $$dW=0$$ is this correct ?

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You should use 1. Conservation of energy and 2. Centripetal force to the problem. I think you can figure out that when the angle subtended by the bead along with x-axis is between 0 and $sin^{-1}(2/3)$, the normal force acting on the bead due to the wire is inwards. Hence, the normal force on the wire would be outwards. It is an easy problem, so I am not ...

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Think about the problem from the point where you neglect the wire and just analyse the motion. You have gravity and at any other point the force has to balance gravity that you end up in a circular motion. In A its towards the origin whereas in B there is no force (except gravity). Different from the classical space-orbit or car-in-a-curve problem is ...

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For a force to do work, it must act upon an object as that object moves some distance (with a component along or opposed to the direction of that force). Holding an object stationary, but carrying the strength of a given force, adds no more work done. Because of this, we say that $$W=\int_C\vec{F}\cdot d\vec{r}$$ where C is the curve in space along which ...

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This is because $P(t)$ as stated is the instantaneous power as a function of time and $W =\mathbf F\Delta \mathbf x$ holds only for constant forces. More generally, recall that a definition of work is the integral: $$W = \int_C\mathbf F(x)\mathrm d\mathbf x$$ Where $C = C(x,t)$ is some curve in space/time. Expressing in terms of time gives: $$W = ... 0 This is not a proper derivation. At a fundamental level, there are at least three important points that are not taken into account by this approach: as you consider a second mass point, it is somewhat difficult to adjust (in a non-arbitrary way) the derivation to obtain the correct energy term related to the angular momentum and/or rigid body rotation ... 2 The dx is actually a vector d\vec{x} (because \vec{x} is actually a vector). A vector's sign depends on your coordinate axes, e.g. if you pick right to be positive then left pointing vectors are negative. In general, the work done W is given by$$ W = \int \vec{F} \cdot d\vec{x} = \int Fdx \cos(\theta) $$where \theta is the angle between ... 0 Pseudoforce are real forces, i.e., they do everything a normal force would do. The problem with conservative nature is that if the observer is moving in a bizzare fashion so that somehow the curl of its acceleration becomes non zero, then the pseudoforce will be non conservative. For observer moving in one direction, force will be conservative and if you ... 0 It all depends on the direction you assign while using dx. It is generally a vector, so the sign will automatically be adjusted according the operation used and its direction relative to other operand vectors. In your case, you are calculating the work done by the spring. The force due to a spring is always directed to its mean position. With this in ... 0 basic concept is to set up a coordinate system , Cartesian would be good here Here, I'm calculating work done BY the spring origin is the point where spring is in natural position (for us to use F = -Kx , otherwise, it would also contain some constant too, ex- if our spring's natural length is at (2,5) and spring only moves in along x axis direction, ... 0 You might get a wrong sign due to cos(0)= 1 and cos(\pi)= -1. You either take x as a cartesian coordinate and integrate over it or you use polar coordinates with some maximum elongation x which is constant. In your question you are mixing up both. The conventional sign here is -, however it depends on your way of counting the work, ie work done to ... 2 The reason is mainly in order to be able to write the total energy of the system as$$ E=T+V $$where T is the kinetic energy of the system. It is far more useful to choose the signs of the terms in the total energy to be +1 once and for all, rather than setting the force equal to plus the gradient of the potential. The link between the conventions is ... 1 How is KE accumulated and thus increasing? This is my primary concern. Also, what's happening with velocity? Is it increasing? Because if KE is increasing then velocity also should increase, right? Between x=0 and x=x_1, a net force, say F, is acting on the body. With Newton's Second Law we get:$$F=ma,$$with a=\frac{dv}{dt} the acceleration ... 2 Because of the Work-Energy Theorem. The work done on an object by an external force \vec{F} is$$W=\vec{F}\cdot \Delta \vec{x},$$where \Delta \vec{x} is the displacement. The Work-Energy theorem says that the work is equal to the change in kinetic energy:$$\Delta K=W.$$So the reason that the kinetic energy is changing for the particle is that ... 1 I think what is missing is the definitions of F and V. Consider first the definition of potential energy. The potential energy at a point relative to another point is the work done by a external force (eg the force exerted by you on the mass, \vec F_{my}) in taking the mass from the first point to the second point. That force which you exert on the ... 3 I think your problem is in how you understand the gradient. Define your z-axis. Lets say it points up. Now$$V(z) = mgz$$. Thus gradient of V(z), that is$$\nabla V(z) = mg \hat z$$. Here,$$\hat z$$is an unit vector pointing up. So the force will be a vector of length mg, pointing down, because of the minus sign. 1 There are some missing words in your description, or the descriptions you've seen. In moving horizontally, gravity does no work because the force and displacement are perpendicular to each other. If you start from rest and apply a force with your hand, then your hand does work as long as your hand is applying a force. Once you reach constant velocity, ... 2 Put on your ice-skate go on some ice-sheet, hold your book and ask a friend to give you a push. Your friend will have to provide some work to make you move but once it is done you are sliding at a constant speed. This is an inertial motion: no force involved no acceleration constant speed. A displacement at constant speed do not require any force, only ... 1 I know this answer is pretty late, but I'm hoping it'll help at least a little. The idea here is, of course, using Green's Theorem:$$\oint \mathbf F \cdot \mathrm{d}\mathbf r = \int_A (\mathbf \nabla \times \mathbf F ) \; \mathrm{d}\mathbf a  Here, you're asking about the line integral around a closed path. When using this equation, you end up making ...

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Quantity that we call work defined as scalar product of the segment of a path and the force applied equals the quantity which we call kinetic energy of an object and/or potential energy of an object. These are all very exact quantities. Furthermore, energy is good for calculations of a motion or predictions of future changes of some variables of a system. So ...

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In another way, we can define work as "the change of energy". When you do work on a system, you change its energy. So positive means you're increasing the system's energy (like pushing a trolley forward), and negative means you're decreasing it. In this context, it might be easier to see why some people are inclined to consider friction as doing "negative ...

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To back up CuriousOne. A good example would be you sumo wrestling with your friend. Lets say your friend is far bigger/stronger than you. So you begin pushing one another. His force on you is greater and so you are obviously moving in what you would call your -x direction. Both of you move in the -x direction, but of course your exerting a force on him too ...

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The question is a bit vague, but the answer is yes. In Thermodynamics and Physics in general, due to the conservation of energy, work is always a transfer of energy between different systems. So we have to talk of what systems we are thinking of. Also, let's start with a simpler situation: consider just flexing your arm through your elbow, just like ...

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Mark said it correctly work is scalar , Moreover , Work is form of Energy . Think of it as you spent some energy on walking and you for sure will not get it back when you stop but instead you'll have to spend some more , the work done basically adds up

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Work is a scalar quantity, which means it doesn't have a direction, unlike velocity or force. So you did work in speeding up and more in slowing down.. I.e the works adds together and doesn't cancel itself out.

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