the child tries hard to push the car with all of her energy but the car does not move at all. In everyday life, that child is said has done a work. This is noticeable because the child looks tired for she has applied a part of her chemical energy. However in physics, that child is said does not apply a work to the car, because the car is not moving. below is given the examples of works that are observable in everyday life.
A child pushing a table, so that is displaces ; a horse pulling cart to the market ; a father pushing a grocery cart down the aisle of a store; an athlete lifting a barbel from the floor to the top of his head, and so on.
In mathematically, work can be expressed by the following equation :
$$W=f .s \cos \theta$$
dengan :W = work (Joule)
F = Force ( Newton)
s = displacement (meter)
$$\theta$$ = angle between force and displacement
The equation of work is used for the force that forms an angle to the displacement. Meanwhile, for the force that is codirection to the displacement, work is determined as follows :
W = F s
in SI, the unit of work is expressed in joule ( J ).
1 Joule = 1 Newton meter
Joule
is used as the name of work unit following the name of james Prescott Joule
(1818 – 1889), an English physicst, and expert in electronics and thermodynamics.
Based on the above relationship of work, force and displacement, we can define
that one joule as the amount of work done by one newton force ( 1 N ) to
displace the position of an object in the same direction with the force as far
as one meter ( 1 m).
Beside
joule, there are other units used to expres work, are erg and calories. Those units
complies with the following relations.
1
erg = 1 dyne centimetre
1
erg = 10^{-7} joule
1
kalori = 4.2 joule
Sample Problem :
Apply
the work equation to determine the amount of work done by the applied force in
each situation described below.
Apply
the work equation to determine the amount of work done by the applied force in
each situation described below.
a.
A
100 N force is applied to a 10 kg object, then the object moves for 5 meters
horizontally.
b.
A
100 N force that forms an angle of 30 degree to the horizontal direction is
applied on a 10 kg object, then the object moves 5 meters horizontally.
c.
An
upward force is applied on a 10 kg object, then the object is lifted for 5
meters.
Solution :
Because
F = 100 N
S
= 5 m
Then,
a.
W
= F s
W
= 100 N x 5 m = 500 J
Thus,
the work done by the 100 N force is 500 J.
b.
$$W=F.s\cos\theta$$
$$W=100x5x\cos{30^{2}}$$
$$W=100x5x\frac{1}{2}\sqrt{3}$$
$$W=250\sqrt{3}$$
W
= 433 J
Thus,
the work done by the 100 N force with the angle 30 degree is 433 J.
c.
W
= F s
F
= m g
W
= m g s
W
= 10 x 10 x 5
W
= 500 J
Thus,
the work done by the upward force is 500 J.
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