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The first law is a statement about the "natural" state of an object. It does not make reference to the specific form of the second law. But keep in mind that the word "force" does appear in the statement (..., unless...). So you question is more about linguistics and mapping English into pure logic than it is about physics. The second law is not needed ...
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No there isn't any problem with applying a force greater than one required to just lift it upwards but then the object would get accelerated and hence would gain some kinetic energy. Usually when we talk about the force applied by the person (lifting the object) it is necessary to know how the force varies with respect to time or position so as to calculate ...
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You are free to exert any magnitude of force you choose and nothing bounds you from doing that.Most of the books use that magnitude of force which is equal to object's weight so that the object is always in equilibrium.In fact most of the books define potential energy in this manner:-
The change in potential energy of a system is defined as the total ...
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Hold a piece of wood against a sanding belt. In your frame, the block is not moving, but
kinetic friction is exerting a force: you have to hold the block still
energy is transferred: the block gets hot, and pieces are pulled off it
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No, this is not always true.
According to conservation of energy, the change in total mechanical energy of a system is determined by the work done by forces external to the system:
$$\Delta E=\Delta K+\Delta U=W_\text{ext}$$
What you are proposing is true if $W_\text{ext}=0$, because then as $\Delta K\geq0$ it must be that $\Delta U\leq0$. Since we started ...
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The key point to avoid confusion is to have clear in mind the correct attribution of forces, work and potential energy to the corresponding physical systems.
When one introduces the potential energy of a body in the gravitational field, the work of interest is the work done by the gravitational force, which, in general is not the same as the work done to ...
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A force is an "instant" phenomenon so to speak. It might depend on time and be applied for a certain period of time. It may do work, such as your airplane thrust, and you can find the energy it transfers as $$W(t_0,t_1)=\int_{t_0}^{t_1} \vec F(t) \cdot \vec v(t) \;dt$$
At each instant, an infinitesimal amount of work $\vec F(t)\cdot \vec v(t)\,\Delta t$ is ...
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Now the work I've done is distance traveled times force magnitude,
hence 20 joules.
That is not strictly true. You are neglecting the kinetic energy that the object possesses at 2 meters due to the velocity you initially gave it has when it reaches 2 meters. The total work you did equals the sum of its kinetic energy (which depends on its velocity) and ...
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The potential
$$U~=~q(\phi - {\bf v}\cdot {\bf A})\tag{1}$$
for the Lorentz force $${\bf F}~=~ q({\bf E} + {\bf v}\times {\bf B}) \tag{2}$$
is a velocity-dependent generalized potential. Such force $\leftrightarrow$ potential interrelation is more subtle:
$$ {\bf F}~=~\frac{d}{dt} \frac{\partial U}{\partial {\bf v}} - \frac{\partial U}{\partial {\bf r}}.\...
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Whenever an object follows a circular path then a centripetal force is required to explain such(the change in direction). The force is given by $F_{centripetal } = m \frac {v^2}{l}$. Therefore here rather than being $T - m g \cos \theta =0$ it is $T - m g \cos \theta =m \frac {v^2}{l}$
[Note: Here $v$ is the instantaneous velocity and varies from one moment ...
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Since $F_{Net}=ma$, if $m=0$ for pulley Q, then you can only have non zero acceleration of the pulley if $F_{Net}=0$ also.
If we can assume the string is ideal (massless and inextensible), and there is no friction between the string and the pulleys, the tension, T, throughout the string must be the same. (I said assume because the complete problem ...
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Yes work done by kinetic friction may be zero for example:- consider a block slipping on ground work done by kinetic friction will be negative in ground frame but now observe the block w.r.t block itself now work done by each and every force will be zero as displacement of block w.r.t itself is zero.
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It's a flat ground, so it can't cause any sideward force. But the water still moves. Whats going on here?
The defining property of a fluid is that the shear stress is proportional to the shear rate. This means that if you have two parallel plates separated by a layer of fluid then the force required to slide the plates (shear stress) is proportional to how ...
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I found explanations of this unsatisfactory as well, including in answers to related questions like this one. Here is a direct explanation in terms of equilibrium of forces.
The green rectangle is the lever, with thickness $h$. (It's upside-down from a typical picture, but that doesn't matter.)
Because the system is in equilibrium, we have $F_x = G_x$ (...
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