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The velocity jacobian is $\vec{\omega}_B = J\, \dot{q}$ with $q=(\phi,\theta,\psi)$. This is used to transform between the generalized forces/torques $Q$ and the vector torques $\vec{M}_B=(\tau_x^B,\tau_y^B,\tau_z^B)$ $$ Q = J^\intercal \vec{M}_B $$ The power through the joint is $$ Q \cdot \dot{q} = Q^\intercal \dot{q} = \left(J^\intercal ...


1

Both can happen simultaneously. This happens in binary star systems all the time. Work done on object A as it moves from position 1 to position 2 is calculated by $$\int_{\vec{r}_{1A}}^{\vec{r}_{2A}} \vec{F}_{total\,on\,A}\cdot\,d\vec{r}$$. That changes the kinetic energy of A. It could be negative which means the kinetic energy decreases. Some might say ...


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Fundamentally you just need to know which direction energy flows. If energy flows from object (or system) A to object (system) B, object A is doing work on object B which is the same as saying object B is being worked on by object A.


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The work done by the painter is $1.93$ kJ, which represents the force he applied to the rope times the length of the pull. The work applied to the barrel is $1.47$ kJ. The rest went into friction. I believe the question is ambiguous between the two values, but that is an English question, not a physics one.


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Because the barrel is moving at constant speed. The tension in the rope is equal to the weight of the barrel. So the answer should be your 2nd value.


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Work is transfer of energy from one system to another OR transformation of energy from one form to another. Either way, work does not create energy. When I lift an object, I am transferring energy from my body/muscles to the object-earth system. The energy goes into potential energy of the object-earth system because the separation between the object and ...


0

Work is defined as $\displaystyle\int F(d) \cdot d$ where $F(d)$ is the net force acting on the object. That's it, that's the definition. You might ask; why do scientists talk about energy and work so much? The answer is that experimental science shows that the energy in a system is constant, and thus energy is a useful concept which can be talked about in ...


0

Net work is basically defined as $\displaystyle\int F(s) \cdot s \; \mathbf{d}s$, where $F(s)$ is the net force – the sum of all forces acting on the body. For example, when a car holds a constant velocity on a highway the net work done is zero (since the net force is zero) but that doesn't mean work isn't done. Both the engine and drag due to air ...


0

The total work done by all forces acting on an object throughout the motion interval of interest is what the work-energy principle involves. It never says "no forces are doing work." And it doesn't talk about changes in potential energy. The changes in potential energy are involved in the work done by the conservative force attached to the particular ...


0

Work is a definition of the expenditure of energy over time. That's it. It's not contained in the definition of any other physical quantity. One could make lots of examples that generate work with chemical reactions (internal combustion engines, explosives), mechanical dynamics (piston firing), electromechanical transformation (electric motor), and so ...


1

When you lift an object (with no increase in kinetic energy), you're doing positive work to lift it up, while gravity is doing negative work (trying to pull it down). In total, the net work is zero. Another way to think about it: total work is net force times displacement. If the object isn't accelerating, you know from $F=ma$ that net force is zero.


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You're completely correct, letting an object fall under the influence of gravity will not change its kinetic+potential energy, it just transforms potential energy into kinetic energy, while leaving the total constant. However, OTHER forces could cause the kinetic energy to increase without changing the potential energy. Imagine a flat ice rink, and a puck ...


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It seems to be your assignment, problem set or so: So instead I will not be spoon feeding you much, here is the method: Use this formula: Work = FxD = KEf - KEi where: KE - Kinetic energy F - Friction D - Distance Remember: KE = mxv/2 Note: Convert units to SI


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A simple solution can be like this- At first you find the rate of change of Kinetic Energy. The process is as follows. $$d/dt(1/2 mv^2)$$ =$$F.v$$ Now imagine this case of a constant force acts on the body which equals to -mg(minus indicates downward direction). So, the equation is -mgv.Now the velocity is the rate of change of vertical position of the ...


1

Yes, if you imagine a projectile shot from ground level, when it hits the ground again the total work done by gravity will be equal to zero. To clear up a possible misconception: When you throw a projectile the initial and final kinetic energies are not zero (otherwise the particle wouldn't have any velocity and so wouldn't move!). However the initial and ...


1

Feynman's trajectory The trajectory discussed by Feynman is shown below in red for the blue path, which is a hyperbolic deflection of a small particle around a large star centered at $(0, 0)$. Discussion Feynman's trajectory here trying to answer the question: how much has the speed increased between A and B. He is answering that by saying that there is ...


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Starting from the work around a close curve $$W = \oint \vec F(r)d\vec s = 0$$ now be stokes throem: $$ \oint \vec F(r)d\vec s = \iint_A \vec \nabla \times\vec F(r)d\vec A = 0$$ where $A$ is the area surrounded by the closed curve. Now if you do the math you will see that $ \vec \nabla \times \vec \nabla U( r) = 0 $ for any field $U( r)$ thus we can ...


1

The reason for this is that, in your notation, $W=C_1-C_2$. This means that the work performed in moving about a circuit from point $A$ to point $B$ along curve $\mathcal C_1$ and then back along $\mathcal C_2$ is exactly the difference between the work performed in moving the particle along $\mathcal C_1$ versus moving it along $\mathcal C_2$. This is ...


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If a mass M is at a point travelling at a given velocity (= speed and direction vector) then it has a given amount of energy (potential and kinetic energy summed) relative to being stationary* at the same point with respect to a given frame of reference. ie solely by knowing position and velocity vector the PE & KE are defined. (* or at some zero ...


0

Here is what I think he means: first we have a planet going around the sun in some orbit, then we change the direction of the velocity to go radially outwards, for example by letting the planet go inside some pipe we put in it's path (notice that a normal planet would never do this, because there are no big pipes in space and also there would be quite a lot ...


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To give you the idea of the complexity of this problem, consider that the dynamics of a single fluid is governed by the Navier-Stokes equation, which is already impractical to solve. $$\frac {\partial}{\partial t} (\rho u) + \nabla \cdot (\rho u \otimes u + p I) = \nabla \cdot \boldsymbol \tau + \rho g$$ Then this equation is coupled to probably three ...


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SECTION A : The example in Feynman's Lectures Let a body P (Planet or Particle or whatever) moving in orbit around a center of attraction called $\:\rm{SUN}$, as in above Figure. Suppose that the attractive force $\:\mathbf{f}\left(r\right)\:$ depends continuously only on the distance $\:r\:$ of the body P from the center $\:\rm{SUN}$. Here it's not ...


3

The point is that if $\frac{1}{2} mv^2 - GMm/r$ is constant, then $v$ only depends on $r$! This is surprising and very useful; it means that $v$ will be the same no matter what path a planet takes from some $r_1$ to $r_2$. In this case, the two paths he's using are the planet's usual elliptical orbit, and a path that goes straight toward the sun. You don't ...


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Apparently, it is assumed that rolling friction can be neglected. In this case, the answer does not depend on the slope of the hill (the kinetic energy does not change, and the change in potential energy only depends on the difference in altitude).


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A conservative force must satisfy the property that the total work done must be independent of the path traveled. In physics, work is defined as the force along a given path times the distance of this path so that, simply: $$ \text{work} = \sum_\text{paths} F\cdot l_\text{path} = \sum_\text{paths} \text{force along path}\times \text{path distance} $$ ...


0

The battery has to do some work in order to create a static electric field inside a capacitor.The current from the battery has to overtake the resistance in order to flow and to reach the charges into the capacitor.So some heat is generated which causes waste to energy.Here the work is done by the battery to charge the capacitor.But the force of the static ...


1

Your answer lies in the ideal gas law: $$PV = nRT$$ where $P$ is pressure, $V$ is volume, $n$ is the amount of substance (usually in moles), $R$ is the "ideal gas" constant, and $T$ is temperature. You can see from the equation that if you're adding substance (i.e. increasing $n$), $V$ must increase proportionally (i.e. the piston must be displaced) to ...


0

In drawing an analogy between electric fields and magnetic field, you forget that there are electric monopoles while there are none such magnetic monopoles. That's why you cannot directly find something similar to the work done on a charge in a capacitor for n inductor. That does not mean magnetic fields don't do work. They do work on magnetic dipoles, ...


1

Your conclusions are pretty much correct. The simplest way to see that magnetic fields don't do work is to consider the $\mathbf F_B = q\ \mathbf v \times \mathbf B$ : that is, the force due to the magnetic field is always perpendicular to motion, so no work is done directly by magnetic field. It's maybe more accurate to say the electromagnetic field can ...


1

you're still converting the same amount of energy to heat, correct? Yes. In other words, brakes made from the same material on the same wheel should experience equivalent wear if each one converts X Joules of kinetic energy to heat energy, no matter if they are applied on the front or back wheel? I don't know. Why do you expect that "wear" is ...


0

Actually, if you apply the front brakes, there is a tendency to topple, while if you apply the rear brakes, there is a tendency to skid. Now, as the linear momentum is converted to angular momentum in this case, by an impulsive brake force, there is some loss of energy. Also, the time of contact of the brakes is less. That is why applying the front brakes ...


3

The answer depends on whether the wheels skid. When you brake with just the rear wheel, it's quite possible to skid; if you apply the front brake, the increase in normal force on that wheel tends to prevent skidding (although in extreme cases it could make you fly over the handlebars). Applying the rear brakes hard enough to block the wheel would generate ...


1

It doesn't matter if the brakes are the same. The bike+rider has a certain kinetic energy. If you stop it, you need to dissipate that much energy. The problem with braking the rear wheel is that braking reduces the load on the rear wheel making it prone to skid. If you brake gently enough, the wheel will not skid. You will then dissipate the kinetic ...


2

In the systems you describe, each string connects always two masses. The tension force exerted on these two masses by the string is equal in magnitude and opposite in direction with respect to the displacement. Hence, the work done by each string on the two masses attached have opposite sign. As a consequence, if you sum up all contributions from each ...


0

First of all, this is called flow work and it is calculated in this way when we have when the system is irreversible in nature when work crosses the boundary when the thermodynamic system is an open system e.g. turbines, compressors, pumps etc etc then we use integral -vdp to calculate the flow Work


0

By saying that it is insulated they are saying that heat can't enter or leave the system. So you have no change in entropy due to $\Delta S = \Delta Q/T$. So if the entropy is constant throughout the process you can use the reversible adiabatic formula. And yes the temperature and pressure do change as you compress the gas (they increase). Strictly speaking ...


0

The general statement of the conservation of the energy is $$ W_{\textrm{non-cons}}= (T_{\textrm{init}} - T_{\textrm{fin}})_{\gamma} $$ hence the work done by any non conservative force (in this case the friction) is equal to the difference in kinetic energies along the path $\gamma$. Friction is a non-conservative force, that is, by definition, there is no ...



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