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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. 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 string and each mass, the net work ...


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 ...


1

Is this the correct way to find the derivative of kinetic energy? $$ K=\frac{1}{2}m v^2 \\ $$ So: $$ \frac{dK}{dt} = \frac{1}{2} \left(\frac{dm}{dt} v^2 + 2mv \frac{dv}{dt} \right) $$ If the mass does not change over the time, then $$\frac{dm}{dt}=0$$ And finally $$ \frac{dK}{dt} = \frac{1}{2} \left(2mv \frac{dv}{dt} \right) $$ So simplifying: $$ ...


3

Here is the procedure: $KE = 0.5mv^2$ $\frac{d}{dt}KE = 0.5m\frac{d}{dt}v^2$ So the question becomes,how do we find the derivative of $v^2$ with respect to time? One can easily see that $\frac{d}{dt} = \frac{dv}{dt}\frac{d}{dv}$ (Notice how the $dv$ cancels top and bottom) Therefore, $\frac{d}{dt}v^2 = \frac{dv}{dt}\frac{d}{dv}v^2 = \frac{dv}{dt}\times ...


2

The time derivative of $v^2$ is $2v \frac{dv}{dt}$ not $2v$. You must use the chain rule.


3

Yes, there are an infinite number of solutions, though your teacher will want you to choose the most obvious one. When the force does work on the mass, that work can be converted into two forms: the potential energy of the object the kinetic energy of the object If you apply a force of $800g$ then once the object has been raised the 2.4m it will still ...


0

When a billiard ball is pushed by a cue, the K.E. of the ball increases as it gains a velocity from rest ( supposing it is at rest). Well, then the internal energy of the ball increases. Thus the work done is done by you on the ball and definitely non-zero.


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The First Law formula you cite is for work done BY a system. If the system is a billiard ball and you hit it with a cue, you do work ON the system. The formula in that case would be delta U = Q + W. The work increases the internal energy of the billiard ball.


0

To speed up the ball, you must move it over a short distance $x$ while applying a force $F$. Thus you add work $W=Fx$. Kinetic energy is internal energy. The work you do raises $\Delta U$ in the shape of kinetic energy $K$. In general the 1st law $\Delta U=Q-W$ consists of: $W$ work: When a force causes motion (like expansion/compression, ...


1

In mechanics, a mass $m$ experiences a force $\textbf{F}$ along some path $C$. The work done on the mass is given by $$ W = \int_C \textbf{F} \cdot d\textbf{r},$$ such that the energy of the mass increases by $W$. Positive work corresponds to energy being added to the system in question (which is inevitably taken from the surroundings). Edit: To answer ...


0

Use the definition of work done over a period of time $t$ , and you have : $$ E_k = \int_0^{t} \vec{F} \vec{dx} = \int_0^{t} \vec{v} d (m\vec{v}) = \int_0^{v} d\bigg{(} \frac{mv^2}{2} \bigg{)} = \frac{mv^2}{2}$$


2

"The answer is that since we are proud physicists and not nitpicking mathematicians we will just wing it when the need arises" This quote is taken from A. Zee's Quantum Field Theory in a Nutshell, and it summarizes the attitude of physicists to mathematics. (At least in an undergraduate level) Since we are physicists, most of our mathematics isn't rigorous. ...


0

The reason for why you do not need to supply any energy is because there is no net work done. Lets assume for a second your container exists in a vacuum. This is essentially an isolated system, not considering black body radiation. This means that none of the energy contained in the box ever escapes and is doomed to stay in the box for all eternity. If ...


0

The real reason is that the gas and container are assumed to be in thermal equilibrium. If the container could radiate into void, then the pressure would slowly decline as the temperature fell. The ideal gas law ignores inter-molecular forces and the finite size of molecules, so eventually the gas will condense and other forces, other than elastic ...


0

Since the scales of the graph are in Newton and meters, your answer is fine. Think of integration as simply adding small individual work for small displacements. For a displacement ds, work done will be dw=F.s Its unit will be in J (or Nm), right. Integrating dw will not change the unit, as adding doesnt change the dimensions. Hope the explanation is ...


1

The slope you calculated is $\frac{10 \rm N}{8\rm m}=1.25\,{\rm \frac Nm}$, so the force function becomes $F=1.25\,{\rm \frac Nm}*x$. Hence, the work is $$ W=\int{(1.25\,{\rm \frac Nm}\times x)dx}=1.25\,{\rm \frac Nm}\int{xdx}=0.625\,{\rm \frac Nm}x^2 $$ (If $F(0)=0$). Because the unit of $x^2$ is $\rm m^2$, then the unit of the work is, as expected, $\rm ...


0

There is no fundamental physics reason why carrying a bag should use more energy than wheeling it. Whether you be carrying a bag or wheeling it, you are essentially sliding it along a line of almost constant gravitational potential (aside from a little jiggling up and down with your stride), so the bag's total energy isn't changing and in theory does not ...


-1

a) You're correct. Since the two carriages AND the spring can only move in one dimention, the system has only one degree of freedom. But wait. They are not moving in a plane! (that is a two domentional space), but in a straight LINE (that has only one dimention). *I know, this is only semantics. b) When you are told that the spring is based on a linear ...


0

If you "push the pedal" (presumably the accelerator) and the car isn't moving, then 2 things are happening. 1) The engine is burning fuel, the pistons are going up and down, and the crankshaft is turning. All of this takes energy. 2) But the wheels aren't moving. This says that something is interrupting the transfer of motion from the crankshaft to the ...


1

(Wheels will not work on a frictionless surface... But let's assume you mean that surface under the box is frictionless and not the surface under the car) then I will have to keep pressing on the pedal of the car so as to apply a force that continuously counteract Something is wrong here. For equilibrium and a non-moving box, you don't want the box or ...


0

I am trying to go to a bit basic level. The formula work=Force*Displacement works only if the force is constant and not changing its direction or magnitude. When an object moves in circle,the force continuously changes its direction. So to calculate it we have to use integral of F with dl,assuming that force remains constant for a very short displacement dl. ...


0

Work is defined as the line integral $\int \mathbf{F} \cdot \mathbf{d\ell}$. The force on an object can be a function of position or time, and could represent external forces placed on the system. Net and total work refer to the same concept, the sum of all work done on an object. For your example, you cannot simply say work is 0 because the object returns ...


1

When you push the block the block 'pushes' you with the same force and you both gain equal and opposite direction momentums. Both block, and you have now some momentum and hence kinetic energy. The work done on both you and the block is: $$ W=\int F \,dx$$ where $F$ is a force applied. It must be equal to the total kinetic energy of you and the block (if ...


0

In the pure hypothetical situation you pose, you, the block, and no forces from gravity, friction or wind resistance - the moment you 'push' the block you will impart a 'packet' of energy for only as long as you can extend your arms and maintain contact with the block. For at that moment the block will move away from you, and you will move away in the ...


2

A mass is attached to a rope, and put into a circular motion. ... I am applying a force only in the radial direction, so how can the tangential velocity increase if there is no tangential force? For short, I'll call the mass attached to the rope a "rock". So how does the rock gain angular velocity? If you truly are applying a purely radial force, and if ...


12

The object will move in a curved path whose center is not where I am pulling from. This center, my hand and the mass form a triangle whose lead angle might be positive or negative depending if the speed of the mass is increasing or decreasing. Consider the body above at B moving along the indicated curved path (like a closing spiral). While pulling from A ...


1

Only a radial force is applied so that angular momentum $L=rmv$ is conserved, $$ dL = m(dr\, v+dv\, r)=0. $$ The force imparts an impulse on the system. So the system is starting to move towards a new equilibrium with shorter radius. The constraint $$ r\, dv = -v\, dr $$ now means that for negative $dr$ the system picks up a positive angular ...


0

I would like to clarify that frictional force is actually doing work, but the work is negative, i.e. energy is transferred away from the object due to the frictional force. With that in mind, for the object to remain its velocity (thus its kinetic energy), work done is required to transfer energy into the object. I think you're right that there is no ...


2

If $\vec{F}$ is a conservative force field, then it satisfies the property $$ \tag{1} \vec{\nabla} \times \vec{F} = 0, $$ and it can be written as $$ \tag{2} \vec{F} = \vec{\nabla}V, $$ for a scalar function $V$ (which corresponds to potential function in physics). Note that, when you put $(2)$ into $(1)$ it becomes a "curl of a gradient" and is ...



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