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Written on 19.13 by sianturikomkom

http://theory.uwinnipeg.ca/physics/work/index.html
http://www.efm.leeds.ac.uk/CIVE/CIVE1140/section03/mechanics_sec03_full_notes02.html
http://hyperphysics.phy-astr.gsu.edu/hbase/work.html
http://www.edinformatics.com/math_science/work_energy_power.html
http://www.sciencejoywagon.com/physicszone/05work-energy/

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concepts

Written on 20.14 by sianturikomkom

Work refers to an activity involving a force and movement in the directon of the force. A force of 20 newtons pushing an object 5 meters in the direction of the force does 100 joules of work. Energy is the capacity for doing work. You must have energy to accomplish work - it is like the "currency" for performing work. To do 100 joules of work, you must expend 100 joules of energy. Power is the rate of doing work or the rate of using energy, which are numerically the same. If you do 100 joules of work in one second (using 100 joules of energy), the power is 100 watts.

Index Work concepts

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Power

Written on 19.48 by sianturikomkom

The quantity work has to do with a force causing a displacement. Work has nothing to do with the amount of time that this force acts to cause the displacement. Sometimes, the work is done very quickly and other times the work is done rather slowly. For example, a rock climber takes an abnormally long time to elevate her body up a few meters along the side of a cliff. On the other hand, a trail hiker (who selects the easier path up the mountain) might elevate her body a few meters in a short amount of time. The two people might do the same amount of work, yet the hiker does the work in considerably less time than the rock climber. The quantity that has to do with the rate at which a certain amount of work is done is known as the power. The hiker has a greater power rating than the rock climber.Power is the rate at which work is done. It is the work/time ratio. Mathematically, it is computed using the following equation.


 
The standard metric unit of power is the Watt. As is implied by the equation for power, a unit of power is equivalent to a unit of work divided by a unit of time. Thus, a Watt is equivalent to a Joule/second. For historical reasons, the horsepower is occasionally used to describe the power delivered by a machine. One horsepower is equivalent to approximately 750 Watts.

Most machines are designed and built to do work on objects. All machines are typically described by a power rating. The power rating indicates the rate at which that machine can do work upon other objects. Thus, the power of a machine is the work/time ratio for that particular machine. A car engine is an example of a machine that is given a power rating. The power rating relates to how rapidly the car can accelerate the car. Suppose that a 40-horsepower engine could accelerate the car from 0 mi/hr to 60 mi/hr in 16 seconds. If this were the case, then a car with four times the horsepower could do the same amount of work in one-fourth the time. That is, a 160-horsepower engine could accelerate the same car from 0 mi/hr to 60 mi/hr in 4 seconds. The point is that for the same amount of work, power and time are inversely proportional. The power equation suggests that a more powerful engine can do the same amount of work in less time.
A person is also a machine that has a power rating. Some people are more power-full than others. That is, some people are capable of doing the same amount of work in less time or more work in the same amount of time. A common physics lab involves quickly climbing a flight of stairs and using mass, height and time information to determine a student's personal power. Despite the diagonal motion along the staircase, it is often assumed that the horizontal motion is constant and all the force from the steps is used to elevate the student upward at a constant speed. Thus, the weight of the student is equal to the force that does the work on the student and the height of the staircase is the upward displacement. Suppose that Ben Pumpiniron elevates his 80-kg body up the 2.0-meter stairwell in 1.8 seconds. If this were the case, then we could calculate Ben's power rating. It can be assumed that Ben must apply an 800-Newton downward force upon the stairs to elevate his body. By so doing, the stairs would push upward on Ben's body with just enough force to lift his body up the stairs. It can also be assumed that the angle between the force of the stairs on Ben and Ben's displacement is 0 degrees. With these two approximations, Ben's power rating could be determined as shown below.


Ben's power rating is 871 Watts. He is quite a horse.

The expression for power is work/time. And since the expression for work is force*displacement, the expression for power can be rewritten as (force*displacement)/time. Since the expression for velocity is displacement/time, the expression for power can be rewritten once more as force*velocity. This is shown below.


 
This new equation for power reveals that a powerful machine is both strong (big force) and fast (big velocity). A powerful car engine is strong and fast. A powerful piece of farm equipment is strong and fast. A powerful weightlifter is strong and fast. A powerful lineman on a football team is strong and fast. A machine that is strong enough to apply a big force to cause a displacement in a s

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Power: Definition, Concept And Example. Energy, Work Output.

Written on 19.07 by sianturikomkom

In daily life, it is interesting, not only to know the work that can be done, but also the speed or rate the work is done.

The work a person can do is limited not only by the total energy output, but also by the speed that the person is able to transform that energy into an specific work output.
Power can be defined either as the rate at which work is done, or, as the speed of energy transfer per unit time.
In the International System, power is expressed as
Joules per second, unit that receives the name
Watt (W), 1 W = 1J/s.
When we say a small bulb wastes 60 watts, we are saying it transforms 60 Joules of electric energy every second into luminous energy and thermal energy.
For large power measurements we use the Horse Power, abbreviated as Hp, equivalent to 746 Watts.
Sometimes it is convenient to express power in terms of the net force F applied to an object and its velocity.

P = W/t, and as W = Fd and v = d/t where d represents displacement, we get,
P = Fv, this is, force times velocity.

Example 1.
Calculate the power a 1200 kg automobile requires for the following situations:

a) The automobile moves up a long hill inclined 8º with a constant velocity of 12 m/s.
b) The automobile accelerates from 14 m/s to 18 m/s in 10 s to pass another vehicle, on an horizontal road. Suppose a constant friction force for both situations equal to Fr = 500 N.

F denotes the force generated by the engine.
SOLUTION.
a) At constant velocity the acceleration is zero, so we can write:

F = Fr+mgsen
F = 500 N+1200 kg·9,8 m/s² ·sen8º = 2.137 N

Using P = Fv, results P = 2.137N·12m/s = 25644 Watts, and when expressed in Hp, it is 34,3 Hp.
b) the acceleration is (18m/s-14m/s)/10s = 0,4 m/s².
By the Second Newton's Law, the sum of all the external forces must be equal to ma, mass times acceleration.

F - Fr = ma
F = 1200kg·0,4m/s² + 500N = 980 N

The necessary power to reach 18 m/s and pass the other vehicle is

P = Fv = 980N·18m/s = 17.640 Watts or 23,6 Hp.

Example 2.
An elevator whose mass is 1,195 kg can carry passengers with up to 935 kg mass. A 4,255 N constant friction force is opposite to the upward movement, as it is shown in the free body diagram below. a) What is the minimum power required by the engine elevator to make sure to lift the elevator at a constant speed of 4.5 m/s? b) What is the required power when the elevator is designed to develop an upward acceleration of 1.75 m/s² and the speed is 4.5 m/s?.
Solution:
a) The free body diagram indicates the engine must develop a force T to rise the elevator. f represents the constant friction force and mg, the elevator weight.

Constant speed means zero acceleration, a = 0.
Second Newton' Law states:
Where m is the total mass equal to 2,130 kg (elevator plus passengers). Then
T = f + mg
T = 4.255x10³ N + (2.13x10³kg)(9.8m/s²) T = 2.51x10⁴N
The power P is:
P = Tv = (2.51x10⁴ N)(4.5 m/s) = 11.3x10⁴ W
b) Now the engine must do additional work for accelerating the elevator. The only change in the setting of the problem is that a 0
F_y = T - f - mg = ma
T = m(a + g) + f
= (2.13x10³kg)(1.75 + 9.8)m/s² + 4,255 N = = 2.89x10⁴ N
The necessary power P is:
P = (2.89x10⁴ N)(4.5 m/s) = 13.0x10⁴ Watts

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Potential Energy - Gravitational and Elastic. Concepts.

Written on 19.06 by sianturikomkom

Potential Energy, Concepts and Example.

An object has energy when it is moving, but it can also have potential energy, which is the energy associated with the object's position.
Potential Energy, Example: a heavy brick lifted up has potential energy due to its position in relation to the ground. It can do work because when dropped it will fall because of the gravity force, allowing it to make a work output over another object receiving the impact.
A compressed spring has potential energy. For instance, the spring of a mechanical clock transforms its energy doing work to move the seconds, minutes and hour pointers.
There are several kinds of potential energy: gravitational, elastic, electric, etc.
Gravitational Potential Energy
The gravitational potential energy is a very common example of potential energy.
The Gravitational Potential Energy (GPE) of an object of mass m at a height y over a reference level is defined as:

EPG = mgy
g is the gravity acceleration

This definition is fully compatible with the definition of work since the work needed lo lift the mass m from the reference level to the height y is Fy = Weight·y = mgy. The object has gained an energy mgy.
If we let this object of mass m to fall freely by gravity on a stake on the ground, the work on the stake will be equal to the kinetic energy acquired while falling.
This kinetic energy can be calculated by the kinematics equation v_f² = v_i² + 2gy. Since v_i = 0, then
vf² = 2gy. The kinetic energy just before striking the stake is ½mv_f². Replacing v_f² with 2gy we get ½ m·2gy = mgy.
Then, to raise an object of mass m to a height y we need a work amount equal to mgy, and once at this height y, the object has the capability of doing work equal to mgy.
Let's notice GPE depends on the object's vertical height over some reference level; in this example, the ground.
The work needed to lift an object does not depend on the lifting path. That direction can be vertical, inclined, or another, and the work to rise the object will be equal.
Also, the work the object is able to do when falling does not depend on its path.
From what level must the height y be measured? What matters here is the potential energy change and we choose a reference level convenient to solve a given problem. Once we choose it, we must keep it during the calculations.
Elastic Potential Energy.
It is the energy associated with elastic materials. Next, we demonstrate the necessary work to compress or stretch a spring over a distance x is ½kx², where k is the spring's constant.
By Hook's Law, the relation between force and displacement on a spring is F = -kx. The minus sign is due to the force always pointing to the equilibrium position (x = 0). The force F is now variable and we can no longer use W = Fdcos.
Let's first find a general relation to calculate the work done by a variable force. We will apply it to our spring.
As F_x is nearly constant in each x, W F_{x x}, the total work can be approximated by the formula
If we make the intervals x even shorter, this is, we make x 0, W tends to a limit, this can be expressed as
This formula represents the integral of the force Fx as a function of x:
And this is the area under the curve F_x(x). Of course, this relation includes the case where F_x = F cos is constant.
Applying that relation to the spring, assuming the spring is placed horizontally and connected to a mass sliding over a smooth surface which is also horizontal, and that the spring is compressed over a distance x_max and then released, the work W done by the spring force between x_i = -x_max and x_f = 0 is:

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Kinetic Energy: Its Relation with Movement and Velocity

Written on 19.03 by sianturikomkom

Kinetic Energy and Movement (Velocity).- To obtain the relation between energy and movement, let's imagine a particle of mass m moving in a straight line with initial velocity V_i . A net constant force F is then applied parallel to its movement, over a distance d. Then, the work done over the particle is W = Fd. As F = ma (a, acceleration) and using the kinetic energy formula V_f² = V_i² + 2ad, where V_f is the final velocity, we get:

W = Fd = mad = m[(V_f² - V_i²) / 2d]d
That is, W = ½mV_f² - ½mV_i²
It is clear we have a difference between final and initial quantities.
The kinetic energy (translational energy) of a particle is defined by physicists as the quantity ½mv² .
Ec = ½mv².
W can also be written
W = Ec
That is, the net work done on an object is equal to the change in his kinetic energy. This result is known as the work-kinetic energy theorem.
Let's notice W is the net work done over the object.
Example. Starting from rest, you push your 1.000 kg car over a 5 meters distance, on an horizontal ground, applying an also horizontal 400 N force. What is the car kinetic energy change?; What is its final velocity at the end of the 5 meters displacement? Disregard any friction force.
Solution. The change in kinetic energy must be equal to the net work done on the car,
W = Fd= (400 N)(5 m) = 2.000 J.
The final velocity is obtained from the equation
W = ½mV_f² - ½mV_i², where V_i = 0.
2.000 J = ( ½ )(1000 kg)V_f², from where V_f = 2 m/s.

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Energy, Work, Power: Concepts.

Written on 19.01 by sianturikomkom

Energy Concepts and Definitions-

Energy is one of the most importants concepts in all physics fields and in other sciences.
The energy has the universal property of being constant. It is not created nor destroyed. Its quantity always remains constant and is never lost; the energy just changes into other forms of energy (heat, light, electricity, etc) with its total quantity balance remaining constant. This explain its importance.
Energy can be defined in a traditional way, although not universally correct, as "the capacity of doing work". This simple definition is not very precise nor valid for all kinds of energy, like energy associated to heat, but it is fully correct for the mechanical energy as described below; that will allow us to understand the close relation between work and energy.
But, what is it understood by work? In daily language has many differents meanings. In Physics has a very precise meaning: to describe what is obtained through the action of a force that moves or displaces over a certain distance.
The work done by a constant force, in magnitude as in direction, is defined as: "the product of the magnitude of displacement with the component of the force parallel to the displacement".
In form of equation:
, , where W donotes work, is the component of the force parallel to the net displacement d.
In a more general form it is written:
W=Fdcos, where F is the magnitude of the constant force, d the object displacement and the angle between the force and the net displacement. Notice that Fcos is precisely the F component parallel to d. Work is measured in Newton meters, unity that receives the name Joule (J) . 1 J = 1 Nm.

Let's see an exercise.
A 40 kg box is dragged 30 m on a horizontal floor, applying a Fp = 100 N exerted by a person. Such force acts doing a 60º angle. The floor exerts a friction force Fr = 20 N. Calculate the work done for each one of these forces Fp, Fr, the weight mg and the normal. Calculate also the net work done on the box.
Solution: There are four forces acting on the box, Fp, Fr, the weight mg and the normal.
The work done by mg and the normal N is zero, because they are perpendicular to displacement(=90º for them).
The work done by Fp is:
Wp = Fp x cos (using x in place of d) = (100 N)(30 m)cos60º = 1500 J.
The work done by the friction force Fr is:
Wr = Fr x cos180º = (20 N)(30 m)(-1) = -600 J.
The angle between Fr and displacement is 180º because they point in opposite directions.
The net work can be calculated in two equivalents ways:

• As the algebraic sum of the work done by each force:
  W_NET = 1500 J +(- 600 J) = 900 J.
• Finding first the net force on the object along the displacement:
  F_(NET)x= Fpcos - Fr
  and then doing
  W_NET = F_(NET)x = (Fpcos - Fr)x
  = (100 Ncos60º - 20 N)(30 m) = 900 J.

A moving object has the capability of doing work, and hence it is said it has energy. For instance, a hammer in movement does work on the nail that strikes. In this example, a moving object exerts a force on a second object and moves it a certain distance.

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