Energy

The transfer of ENERGY


Energy has different meanings. The nontechnical usage derives from the Greek en(which means in) and ergon(which means work). Hence energy is the capacity to do work, an inherent vigor.
Energy describes the state of a system in relation to the action of the Four force.
We buy electric energy packageed from the hardware store and on tap from the power or energy company.


Such as, Chemical energy can be gotten from a biodiesel or gasoline.Electrical energy from solar cell. The most important characteristic of energy.Energy transferred from one configuration of interactions to another, such that the total amount of energy remain unchanged. Themal energy can be converted into electrical energy, and some of that turned into light, and back again into thermal energy, but the net amount of energy is always the same.Energy is conserved.

Energy is a scalar quantity associated in various amount with all the things that exist, from minute massless particles to immense whirling galaxies. By observaing the changing behavior of matter, we infer the presence of one form or another of energy. Like linear and angular momentum, energy is not an entity in and of itself.There is no such thing as pure energy.

NEXT : WORK

WORK

When a body has a displacement with magnitude s along a straight line while a constant force with magnitude F, direct along the same line. We define the work W done by the force on the body as


W=Fs................(1)
example 1

Steve is trying to impress Elaine with his new car, but the engine dies in the middle of an intersection. While Elaine steers, Styeve pushes the car 19 m to clear the intersection. If he pushes in the direction of motion with a constant force of 210 N (about 47 lb), how much work does he do on the car?

Solution
From W=Fs
=(210N)(19m)
=4.0 x 10^3 J
Answer.

Steve pushed the car in the direction he wanted it to go. What if he had pushed at an angle phi with the car displacement. Only the component in the direction of the car's motion, (210N)cos(phi), would be effective in moving the car. When the force F and the displacement s have different directions, we take the component of F in the dirction of the displacement s, and we define the work as the product of this component and the magnitude of the displacement. The component of F in the direction of s is Fcos(phi), so

W=(Fcos(phi))s.................(2)
or
W= F.s ........................(3)



brought to you by: University Physics ,Hugh D.Young Eighth edition


Example 2.

Consider a body with mass 0.2 kg move toward the right along smooth horizontal surface by a horizontal force F. The body has initial velocity of 2.8 m/s.
As a body move ,It slows down because of the horizontal friction force exerted on it by the horizontal surface.The displacement of a body is 1.0 m before coming to rest. What are the magnitude and direction of the friction force acting on it?

Solution
From the second law of motion with ay =0 we obtain

Fy=m.ay
or
N-W =0
when N is normal force, W is weight of a body

Hence N=W=mg=(0.2kg)(9.8 meters/sec^2)= 2 N


Suppose the body move toward in the +x direction, starting at the point x0=0 with the given initial velocity. We assume that the friction force f is constant. The acceleration is then constant also.Then we have

v^2=v0^2+2a(x-x0)
0=(2.8 m/s)^2+(2a)(1.0m),
a=-3.9 m/s^2)

The negative sign means that the acceleration is toward the left.
The friction force f then become

f=ma=(0.2kg)(-3.9 m/s^2)= -0.8 kg.m/s^2 =-0.8 N

Answer.


Example 3.
A block of mass 10.0 kg is to be raised from the bottom to the top of an incline 5.00 meters long and 3.00 meters off the ground at the top. Assuming friction less surface, how much work must be done by a force parallel to the incline pushing the block up at constant speed?


Solution

Define F = P= The magnitude of force pushing the block up the incline. Because the motion is not accelerated, the resultant force parallel to the plane must be zero. Thus

P-mgsin(theta) = 0,
or P=mgsin(theta)=(10.0 kg)(9.80 meters/sec^2)(3/5)= 58.8 N

When define s=d =displacement of a block.
Then the work done by P ,theta = 0 degree is

W=P.s=P.d =Pdcos(0)=Pd= (58.8 N)(5.00 meters)= 294 joules.






Answer.


Next: work done by variable force

Work done by a variable force

When the force exerts to the body with force not constant. We can find the work done by integral of force F with respect to displacement x ,that a body move from point x1 to x2

For the stretched spring. To keep a spring stretched an amount x beyond its unstretched length, we have to apply a force with magnitude F at each end. If the elongation is not too great, we find that F is directly proportional to x



F = kx.........................(4)


where k is a constant called the force constant or spring constant of spring.

equation (4) is known as Hooke's law. And the total work done by this force is


W=0.5kx^2

Suppose the spring is stretched a distance x1 at the start. then the work to do it to stretch it to a greater elongation x2 is


W = 0.5kx2^2-0.5kx1^2 ................(5)


NEXT : Work and Kinetic Energy

Work and Kinetic Energy

A particle with mass m moving in a straight line under the action of a constant resultant force with magnitude F direction same along axis . the particle's acceleration is constant . Work done by resultant force F is equal to the total work Wtot done by all the force.
The kinetic energy K of the particle defined as


K= 1/2mv^2

And We can say that the work done by the resultant external force on a particle is equal to the change in kinetic energy of particle

Wtot = K2-K1 = delta K

This result is called the work-energy theorem. Kinetic energy is a scalar quantity.


Example 1. Assume the force of gravity to be constant for small distance above the surface of the earth. A body is dropped from rest at the height h a bove the earth's surface. What will its kinetic energy be just beforce it strikes the ground ?

solution
The gain in kinetic energy is equal to the work done by the resultant force, which here is the force of gravity. This force is constant and directed along the line of motion, so that the work done by gravity is



W=F.d =mgh

Initially the body has speed v0=0 and finally a speed v. The gain in kinetic energy of the body is

1/2 mv^2-1/2mv0^2 = 1/2mv^2 - 0.
Equating these two equivalent terms we obtain

K =1/2 mv^2 =mgh
as the kinetic energy of the body just before it strikes the ground.
The speed of the body is then

v=sqrt(2gh)

If the body is falled from a height h1 to a height h2 a body will increase its kinetic energy from 1/2 mv2^2 -1/2 mv1^2 = mg(h1-h2).

NEXT : Potential Energy And Conservative Forces

Potential Energy and Conservative Forces

An energy associated with the position of bodies in a system. Changes in this energy may accompany opposite changes in kinetic energy in such a way that the total energy remains constant, or is conserved. Energy associated with position is called potential energy, and forces that can be associated with a potential energy are called conservative forces. A system in which the total mechanical energy, kinetic and potential, is constant, is called a conservative system


Example 1.

When we throw a 0.150-kg baseball straight up in the air, The initial upward velocity is 20.0 m/s. Use conservation of energy to find how high it goes, ignoring air resistance.


Solution

the only force doing work on the ball after it leaves your hand is its weight. Let's take the origin at starting point(point1), where the ball leaves your hand; then y1 =0. At this point, v1= 20.0 m/s. We want to find the height at point 2, where it stops and begins to fall back to earth. At this point, v2=0 and y2 is unknown.

Then K1+U1 = K2+U2, or

1/2 mv1^2+ mgy1=1/2 mv2^2+ mgy2,

1/2(0.150kg)(20.0 m/s)^2 + (0.150 kg)(9.80 m/s^2)(0)
= 1/2(0.150kg)(0)^2 + (0.150 kg)(9.80 m/s^2)y2,

y2 = 20.4 m


Answer

Elastic Potential Energy

When a railroad car runs into a spring bumber at the end of the track, the spring is compressed as the car is brought to a stop. If there is no friction, the bumper springs back, and the car rolls away in the opposite direction. During the interaction with the spring, the car's kinetic energy has been " stored " in the elastic deformation of the spring.

The work must do on the spring to stretch it from an elongation x1 to a greater elongation x2 is

1/2kx2^2-1/2kx1^2.


Now we need to find the work done by the spring on the body in a displacement from x1 to x2 the spring does an amount of work Wel given by




Wel =1/2kx1^2 -1/2kx2^2.


and elastic potential energy as
U = 1/2kx^2

Wel=U1-U2=-delta U.

Inelastic collisions

If the interaction force between the bodies are conservative, the total kinetic energy of the system is the same after the collision as before. Such a collision is called an elastic collision.

A collision in which the total kinetic energy after the collision is less than that before the collision is called an inelastic collision. In one kind of inelastic collision the colliding bodies stick together and move as one body after the collision; this is often called a completely inelastic collision.

A completely inelastic collision of two bodies (A and B). Because they stick together after the collision, their final veloxcities must be equal :


vA2=vB2=v2

conservation of momentum gives the relation

mAvA1 + mBvB1 = (mA+mB)v2.

ifd we know the masses and initial velocities, we can compute the common final velocity v2.

suppose, for example, that a body with mass mA and initial velocity v1 along the +x-axis collides inelastically with a body with mass mB that is initially at rest (vB1=0). the common x-component of velocity v2 of both bodies after collision is

v2 =(mA/(mA+mB))v1

Form The kinetic energy K1 and K2 before and after the collision, respectively, are


K1 =1/2 mAv1^2,
K2 = 1/2(mA+mB)v2^2= 1/2 (mA+mB)(mA/(mA+mB))^2v1^2.
The ratio of final to initial kinetic energy is

K2/K1 =mA/(mA+mB)

Elastic collision

From Wikipedia, the free encyclopedia
An elastic collision is a collision in which the total kinetic energy of the colliding bodies after collision is equal to their total kinetic energy before collision. Elastic collisions occur only if there is no conversion of kinetic energy into other forms.

Consider two particles A and B had elastic collision. Let the x component of velocity before the collision vA1 and vB1 and those after the collision vA2 and vB2.

From conservation of kinetic energy.

Total kinetic energy is the same before and after the collision, hence:

1/2 mAvA1^2 +1/2 mBvB1^2=1/2mAvA2^2+1/2 mBvB2^2,

and conservationb of momemtum gives

mAvA1+mBvB1=mAvA2+mBvB2

if the masses mA and mB and the initial velocities vA1 and vB1 are known, these two equations can be solved simultaneously to find the two final velocities vA2 and vB2.

Energy: WORK#links

Energy: WORK#links