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
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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
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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
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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
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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
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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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