tugas 1 work

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WORK AND KINETIC ENERGY

1. The Definition of Work

The concept of work in physics has a particular and has much more

specific definition rather than it commonly used in daily language. In our

daily language the term work is related to expenditure of muscular effort,

while in physic work is defined as force act on a particle and causes it

displaced. Both of this definition does not contradictive to each other, but

there are several cases of work in a term of daily language is not accepted to

be work in physics point of view. Suppose a cafeteria tray at a constant speed

across a table its acceleration is zero and therefore its force is zero. In daily

language he was said to do a work because he gets exhausted while in physics

he or she is not since the displacement is not causes by the applied force. The

unit of work is the unit work done by a unit force in displacing a particle a

unit distance in the direction of the force. In the international system of unit

work is expressed in a unit of Newton-meter which is called Joule.

2. Work Done by a Constant Force

Consider a particle acted on by a force. In the simplest case the force is

constant and the motion takes displacement in a straight line. In this situation

we define the work done by the force on the particle as the product of the

magnitude of the Force and the distance through the particle moves.

Mathematically it can be expressed by:

Equation 1. work is the multiplication

of applied force To the objects displacement



On the other situation the constant force itself acting on a particle may not act

in a direction in which the particle moves. For example suppose a constant

force F makes an angle with the x axis acts on a particle which

displacement is d along the x axis. If W represents the work done by F during

this displacement, then according to the definition

Work is a scalar quantity, even though the two quantities involved in its

definition force and displacement are vectors. Because it is being the dot

product of the multiplication of those two vectors involved. Work can be either

positive or negative. If the particle on which a force acts ha component of

motion opposite to the direction of the force the work done is negative. This

corresponds to an obtuse angle between the force and the displacement vectors.

For example when a person trying to push a book selves on a rough floor and

cause it displace o the same direction with the applied force given. The work

done by the person can be treated as the positive work since the force exerted

causes the object displace to the same direction, but the work done by the

friction between the object and the rough floor is considered to be negative

since the friction force has component of motion opposite to the displacement

of the object.

Figure 2. a force applied on an angle θ to the displacement of the object

Equation 2. work done by force

With an angle θ to its displacement



3. Kinetic Energy and Work-Energy Theorem

On the previous discussion of the work, we only focused at the work on

which no acceleration are applied on it. Let now we moving on to consider

the work on an accelerated object. The simplest situation to consider is that

of a constant resultant force. Such force F acting on an object which has mass

m. according to Newton’s second law, the object will be accelerated by a

constant acceleration a. Let us choose the x axis as the direction of F and a.

how to calculate the work done by this force on this object (suppose it cause

displacement x)?

On a constant acceleration we have relation that acceleration is the rate of

change in velocity within time t . Thus

And the relation between displacement x and the average velocity

m m

x

Figure 2. a block undergoing an acceleration along distance x



Here v0 is the body’s velocity at t = 0 and v is the body’s velocity at the time

t . then the work done is

W =

W = ( ) (

)

W =

We call the one half the product of the mass of a body and the square of it

speed as the kinetic energy of the body. Or on the other word we can say that

the work done is equal to the change of the kinetic energy of the body.

Sample problem:

A cart full of sands on which it mass equals 12 kg at the first

observation is pushed from rest with a constant force 20 N on the frictionless

floor to the right direction. If there is a small hole at the bottom of the cart

which then causes the sand drops 0.01 kg each second, if the mass of the cart

is 2 kg Calculate the total work done by the constant force at time 10 s, 20 s,

30 s, and 40 s.

Equation 3work as the change

Of kinetic energy



Solution :

On this case we have to note that because the mass of the system is

decreasing within time meanwhile the exerted force is constant, according to

the Newton second law about motion, the system has to gain acceleration

within time. To calculate the work done we have to equalize as the change of

kinetic energy on which the initial velocity is 0 m/s therefore

To calculate the magnitude of v on which the initial velocity is zero we can

use the equation

To convenience our calculation the results is shown by the table below

No t (s) F (N) a (m/s2) m (kg) v (m/s) W (Joule)

1 0 20 0 12 0 0

2 20 20 1.69 11.8 33.8 6740.4

3 40 20 1.72 11.6 68.8 27453.9

4 60 20 1.75 11.4 105.0 62842.5

Table 1 the change in mass within time corresponds to the magnitude of the work

done by the constant force



REFERENCES

http://www.cliffsnotes.com/study_guide/Work-and-Energy.topicArticleId

Resnick, Robert and Halliday, David (1960 ) Physics: Parts I and II. New York,

London Sydney: John Wiley and Sons,Inc




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