I work and energy – digital pulse oximeter
Posted: November 3, 2010 at 9:38 am
1 – it criticizes the following definition of energy: “energy is the capacity to produce work.”
Commenting on
IEnergy is a concept difficult to define, although it is family to all. A lot of times we are picked in circular definitions of work and energy, seemingly using the two have to describe the same thing.f “energy is the capacity to produce work”, we asked, then: what is work? Will it be that you will be satisfied with the mathematical expression that quantifies the work?
We don’t know how to define energy, time, space, matter etc. They are primitive concepts. The science has their primitive concepts; the mathematics also has them, for instance, in the geometry they are the point concepts, straight line and plan. In the chemistry they are the one of energy, it structures and connection. Unlike other physical greatness as speed and force, energy is an abstract concept that possesses monetary value when he/she becomes work. A force of intensity 1 kgf or a speed of measure 3 m/s are not things that one can buy or to sell; all of the months, however, we paid to the distributing companies for the kilowatt-hour that we consumed.
2 – being exercised a constant force along a given distance on bodies initially in rest, which will the kinetic energy be at the end of the course? Will the kinetic energy depend on the mass of the body?
Solution
The kinetic energy, at the end of the course, will be the same for all of the bodies.
Let us consider the movement unidimensional of a pot of steep dry (reduced attrition to the minimum) ice for a constant force along a horizontal track of polished aluminum
As the force is constant, the pot will accelerate evenly, with the = F/m and we will have:
x1 = (1/2).(F/m).t2 and v1 = (F/m).t
Being eliminated t among these two expressions, it comes:
v1 = (2Fx1/m) 1/2 and Q1 = m.v1 = (2mFx1) 1/2.
The amount of movement supplied to the pot is not independent of the mass of the pot, as it would happen if we applied the force during a die interval of time. The amount of movement is proportional to the square root of the mass m.
He/she exists, however, an amount that is invariante for all of the pots, independently of their masses, to know:
Ecin. = Q21 / (2m) = (m.v21) /2 = F.x1
Like this, when a constant force is exercised initially along a given distance on bodies in rest, all acquire the kinetic energy, Ecin., in the end of the course, that is, DEcin. = Ecin. = F.x1 = WF, being WF the work accomplished by the force F in the displacement x1.
3 – a boy on pots of dry ice meets of foot on a horizontal surface of polished aluminum, close to a vertical wall. Pushing the wall, the boy goes out with a certain speed and kinetic energy. Who paid for that energy?
Solution
The external forces that act on the healthy boy: the) the weight applied P for the Earth; b) the force N upward applied for the surface; c) the horizontal force F exercised by the wall on the boy’s hand.
Who “paid” for the kinetic energy was the internal work of the boy’s muscular forces. Analytically, for the theorem kinetic work-energy, we have:
Wres. = Wext. Wint. = DW
Now, Wext. = 0, then, Wint. = (1/2) m.v2.
There is another way to face the problem.
The impulsive force, F, acting in the boy’s hand during the interval Dt, it supplies a digital pulse oximeter F.Dt = m.Dv. For other part, the force F moves his/her point of application of Dx and it accomplishes the work FDx = (1/2) mv2, being Dx the small deformation of the boy’s hand.
4 – a person carrying a sack in the head for a horizontal highway accomplishes or don’t I work physical? To discuss.
Solution
In first place, we needed to know the physical work is external or internal. In the case of the physical work to be external, we have to complete the sentence saying: do I work that forces?
In the given case, to I force-weigh doesn’t work, once there is no force-weight component in the direction of the displacement; however, we should not forget the force of attrit
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