Hypothesis
As said the above formula of the weight of an object, the weight equals it
mass multiplied by acceleration (which is always the same quantity, 9.8). What
I am going to do in this work is how changing the surface area of an object,
can affect the velocity of the same. The object is going to be a square of clay
which is going to have the same weight, but I am going to change the surface
area of the square by using the formula of this object (A = l2),
also, the height from which I’m going to throw it is not going to change,
always is 1.75 m of altitude.
What I think is going to happen is that if there is a change on the surface
area, if there is an increase on its surface area, the square would reach more
slowly the floor.
On this experiment, the formula more indicated to see the change is the one
of velocity = distance/time.
LAB REPORT CHANGE IN SURFACE AREA OF AN SQUARE OF CLAY
Height from
where we are going to throw the ball of clay: 1,75 cm
Area of the
square:
Mass with 25m2:
115,66 g
5 cm is the
length of one side of the square
A = 5 * 5 = 25
m2
|
Number of the experiment
|
Surface area
|
Time (s)
|
|
1
|
25 m2
|
0.50 s
|
|
2
|
25 m2
|
0.62 s
|
|
3
|
25 m2
|
0.59 s
|
|
4
|
25 m2
|
0.63 s
|
|
5
|
25 m2
|
0.59 s
|
Mass with 36m2 = 113, 15 g
6 cm is the
length of one side of the square
A = 6 * 6 = 36
m2
|
Number of the experiment
|
Surface area
|
Time (s)
|
|
1
|
36 m2
|
0.66 s
|
|
2
|
36 m2
|
0.62 s
|
|
3
|
36 m2
|
0.62 s
|
|
4
|
36 m2
|
0.59 s
|
|
5
|
36 m2
|
0.60 s
|
Mass with 49m2: 112,13 g
7 cm is the
length of one side of the square
A = 7 * 7 = 49
m2
|
Number of the experiment
|
Surface area
|
Time (s)
|
|
1
|
49 m2
|
0.63 s
|
|
2
|
49 m2
|
0.62 s
|
|
3
|
49 m2
|
0.50 s
|
|
4
|
49 m2
|
0.56 s
|
|
5
|
49 m2
|
0.60 s
|
Mass with 64m2: 113,13 g
8 cm is the
length of one side of the square
A = 8 * 8 = 64
m2
|
Number of the experiment
|
Surface area
|
Time (s)
|
|
1
|
64 m2
|
73 s
|
|
2
|
64 m2
|
70 s
|
|
3
|
64 m2
|
72 s
|
|
4
|
64 m2
|
60 s
|
|
5
|
64 m2
|
62 s
|
Average of the
time
|
Length of the side of the
square
|
Average of the time
|
|
5 cm
|
0.59 s
|
|
6 cm
|
0.62 s
|
|
7 cm
|
0.58 s
|
|
8 cm
|
0.67 s
|
Conclusion:
As we can
observe in our graph and table, our results weren’t conclusive. Our result
weren’t the same as we expected them to be in the hypothesis, and this may be
because an experimental change or fault in our experiment such as not the exact
precision.
The time had to
decrease as the length of the side of the square increase, and it didn’t:
first, it increased, then it decreased and finally the time increased again.
“What I think is
going to happen is that if there is a change on the surface area, if there is
an increase on its surface area, the square would reach more slowly the floor”.
This is what we
supposed in the hypothesis but what finally did happened was different:
What finally
happened in our experiment is that as there is an increase on the surface area,
the square reaches the floor first, quicker, then slower, and finally quicker.
The right result should have
been equal to the hypothesis.
Equations
related with the motion in the air of an object:
V= D/T
Velocity (m/seconds or
km/hours) = distance (m)/time (seconds)
If we are throwing an object
at a high altitude, obviously, it would reach the floor slowly than if the
altitude is smaller, and also, depending on how is the surface area or the
shape of the object, the velocity upon the object would be quicker or not.
W= m*g
Weight of an object = mass of
the object x acceleration of the gravity
If the mass of an object is
bigger, the acceleration of the object is going to be slower and the weight of
the object would be as well as its mass bigger.
Ec= 1/2 m*v^2
Kinetic energy = ½ x mass x
volume square
If there is in an object which
has a volume square and mass is bigger and has a higher quantity, the kinetic
energy of the same, would have an increase.
F= m*a
Force of an object = mass of
the object x acceleration
Frictional force is any force
opposite to a movement, which manifests itself in the contact surface of two
bodies that always one moves or tends to move over another.
If there is an object that has
a big mass and acceleration, its force would increase also, as it depends on
this two factors to show its force.
Fair= -cv^2
The air resistance = air
constant x object’s velocity (square)
Air Resistance Formula is
used to find the air resistance, air constant and velocity of a body if some of
these quantities are known. This is the formula more appropriate for my
experiment.
These are some formulas that helped
us in our project.
Evaluation:
Problem:
Precision. Maybe we should have been more cautious at the time we were doing
the experiment because there are several factors that could have affect the
results:
Not measuring the time exactly. To be more precise with time, we might have
film the experiment so that way we were more precise.
Not throwing the clay with the same force: Maybe we added some force at a
specific time when we were throwing the clay to the floor.
I know that we aren't adding force because we are just dropping the plasticene but perhaps we are pushing the clay a little when we let it drop, this could be prevented just being precise and standing still without moving.
I know that we aren't adding force because we are just dropping the plasticene but perhaps we are pushing the clay a little when we let it drop, this could be prevented just being precise and standing still without moving.
Solution: The
solution to these problems is to be very careful at the time we are doing an
experiment which can be easily affected for small faults
Problem: The
clay had not the same weight always. We used the same piece of clay every time.
So, maybe, at the time it reached the floor, it was losing some its weight
leaving some clay in the floor. And that could have affected the result.
Solution: Making
one piece of clay for each experiment and making sure they al have the
same weight.
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