.

1. PRACTICUM FINAL REPORT
BASIC PHYSICS
"CINETIC AND STATIC FRICTION STYLES"
COLLECTION DATE : 06 NOVEMBER 2017
PRACTICUM DATE : 01 NOVEMBER 2017
A. PRACTICUM OBJECTIVES.
1. Can know the static and kinetic friction coefficient.
2. Can understand the concept of static and kinetic friction coefficients.
3. Can know the acceleration of a moving object.
4. Can know the concepts of Newton's Law.
5. Can know and understand the concept of dynamics.
B. BASIC THEORY.
Intuitively, we experience force as a form of encouragement or attraction to
things. If you push a broken car or a shopping trolley in the supermarket, you direct the
force to the object. When the motorbike lifts the elevator, or the hammer hits the nail,
or the wind blows the leaves in the tree, the force is being deployed. We often refer to
this as a contact force because the force is applied when an object is in contact with
another object. On the other hand, we say that an object falls because of the force of
gravity, which is not a contact force. If an object is in a state of silence, to make it start
moving it takes force, a force is needed to accelerate an object from zero speed to
nonzero speed. For an object that is already moving, if we want to change its speed -
both direction and magnitude - again a force is needed. In other words, to speed up an
object, style is always needed. (Giancoli. 2014: 93).
Physics also learns what causes an object to accelerate. The reason is the force,
which in its free language is a push or pulls speeds up, the force of the track works on
the rear tire of the car and causes the racing car to accelerate. When a defender destroys
the attacker (in American football), the style of the defender works on the attacker to
produce a backward acceleration. When a car hits a telephone pole, the force acting on
the car from the pole causes the car to stop. Science, engineering, legal, and medical
journals are filled with style articles on objects, including humans.
The relationship between the force and acceleration that it produces, was first
understood by Isaac Newton (1642-1727). The study of this relationship, as presented

2. by Newton, is called Newtonian mechanics. Newtonian Mechanics does not apply to
all situations. If the acceleration of objects interacting is very large - for example,
approaching the speed of light - we must replace Newtonian mechanics with Einstein's
special theory of relativity, which applies to all speeds, including speeds close to the
speed of light. If objects interact with atomic-scale structures (such as electrons in
atoms), we must replace Newtonian mechanics with quantum mechanics. Physicists
now see Newtonian mechanics as a special case compared to the two more
comprehensive theories. But Newton's mechanics is still a special case that is very
important to study because it applies to the motion of objects ranging from very small
sizes (almost on the scale of atomic structures) to the size of astronomy (galaxies and
galaxy clusters). (Halliday et al. 2005: 97).
Dynamics studies the motion of objects by observing the causes of motion. An
object that is initially silent, can move due to the force acting on the object. Branch of
physics that studies motion by calculating the forces acting on the object. Is dynamics.
We often encounter dynamic object observations in everyday experience. Why are your
feet more painful when hit by a large stone than a pebble? Why can't we walk in a very
slippery place? When you stand in a city that is running and then brakes suddenly, what
happens? Why is that? These questions are a few simple dynamic examples.
1. Newton's first law.
Basically, every object tends to maintain its position (inert). If the object is idle,
the object will remain silent. When an object is moving, the object will still
move.
The object will remain in its position if there is no external force acting on it.
This view was examined by Newton to produce a concept of the style known as
Newton's first law. Newton's first law states the following: "if the resultant force
acting on an object is zero. Then the object will stay still or keep moving straight
irregularly. (Ruwanto. 2006: 35).
Systematically Newton's first law was written with equations:
Based on these equations, stationary objects and irregularly moving objects do
not experience the resultant force. As a result, the object does not experience
acceleration (acceleration is zero). Therefore maintaining the initial condition
of the object. Newton's first law is often called Law / Inertia. (Intan. 2001: 9)
2. Newton's second law.

3. What happens if the resultant force on an object is not equal to zero? Newton
predicts that if the resultant force is not equal to zero, the speed of the object
will change. The speed of the object will increase if the resultant force is in the
direction of the speed of the object. Conversely, the speed of the object will
decrease if the force of the force acting on the object is the opposite of its speed.
Newton states that if there is an external force acting on an object, the object
will experience an acceleration whose direction is the same as the resultant
direction of that force.
Based on experimental data Newton concluded as follows:
a. The magnitude of acceleration caused by the resultant force acting on
the mass of the object m is proportional to the magnitude of the resultant force.
That is, the greater the force, the greater the acceleration of the object.
b. Object acceleration (a) due to the resultant F force on the object is
inversely proportional to its mass (m). the greater the mass of the object, the
smaller the acceleration of the object for the fixed force F.
From these conclusions formulated Newton's second law which can be
mathematically written as:
3. Newton's third law.
For example, the child pulls the end of the rope in the F style. At the end of the
rope that is attached to the wall, the force of F is given to the wall. In the hands
of the children work the F style given by the rope, as well as on the end of the
rope that is attached to the wall, the working force F 'is given by the wall. As a
result, at the end of the rope works two styles that are equally large and in the
opposite direction that give the resultant noil, so that the rope does not move
and looks tense.
If the force moving on a wall is called the action force, the resistance force (by
the wall) that moves the child towards the wall is called the reaction force. So:
(the negative sign shows the opposite direction)
Faksi = -Freaksi

4. Or, in another sentence states "if an object (first object) works on another object
(second object) then the second object works the force on the first object, equal
to and opposite to the force on the first object. The things that have been
observed above are known as the third of Newton's Law.
The concept of Action-Reaction: The
1. style style of action and reaction is equally large but in the opposite
direction.
2. The pair of action-reaction forces exists if there are two objects interacting.
3. The action force and reaction work on two different objects.
(Hari Subagya. 2012: 88-89).
The friction force on the car tires is made rather rough and structured so that the
frictional forces with the road surface remain. This is intended to avoid slippage. On
the other hand, the friction between the inside of the engine of the car is actually
minimized by using (useful) and some are harmful (need to be avoided or minimized).
Friction style Friction
force is one of the touch styles. The force works on two surfaces of solids that
come in physical contact. There are two types of force, namely static friction force and
kinetic friction force. Static friction forces tend to maintain a state of rest on objects,
while kinetic friction tends to maintain the motion of objects. (Ruwanto. 2006: 36-37).
Advantages of friction include:
1. Use of nuts and bolts as a binding device.
2. Brakes on the wheels of the vehicle so the wheels don't slip.
3. The atmosphere of the earth protects it from meteors. Because it gets friction from
the atmosphere, meteors that fall from space burn before they get to earth.
The disadvantages of friction include:
1. Accelerating vehicle tires to wear out.
2. Air friction slows the plane.
3. Making motor wheel heat.
To avoid these losses, we must minimize friction. Ways to minimize friction: copy
the surface, use wheels, reduce the surface. (Rahayu. 2013: 109)
Static friction force.
The static friction force is the friction force that occurs between two surface
objects that are stationary or there is no relative motion of one to the other. The motion

5. of an object even though there is a force acting on the object indicates that the resultant
force acting on it is zero.
You try to launch a box of books on the floor, but the box doesn't move at all.
That is because the pull or push from you is equal to or smaller than the other forces
acting on it, but in the opposite direction. The style is called static friction with the
symbol fs.
Static friction force has a value that lies between zero to a maximum value of
μs N,
(Ruwanto. 2006: 37)
Kinetic
friction force Kinetic friction force, the force required to maintain the motion of
an object smaller than the force needed to start moving the object. To start moving an
object, the external force is first used to overcome the static friction force of the object
with the other surface that touches it. After moving, some of the external force that
maintains the motion of the object is used to overcome the kinetic friction force, i.e.
The magnitude of the kinetic friction is smaller than the maximum static friction force,
where fk <fs max.
Kinetic friction reflects the relative relationship between the two surfaces that make
contact. The magnitude of the kinetic friction is:
With μk, the kinetic friction coefficient is called. (Ruwanto. 2006: 38).
C. TOOLS AND MATERIALS
NO IMAGES
NAME OF TOOLS AND
MATERIALS
1 Ruler

6. 2 Beam
3 Stopwatch
4 Digital Balance
5 Inclination board thanks to
NO PICTURE NAME OF TOOLS AND MATERIALS
6 Nylon Thread

7. 7 Inclination load restraint.
D. WORK STEPS (STATIC STYLE STYLES)
LICIN SURFACE.
NO PICTURE
OF NAME OF EQUIPMENT
AND MATERIALS
1
Prepare a tool for practicing
static and kinetic friction forces.
2
Weigh the mass of the slippery
surface.
3
Place the inclination board on the
table.
NO IMAGES
OF NAMES TOOLS AND
MATERIALS

8. 4
Lift the inclination board with an
inclination load brace and place
the slippery field beams on the
inclination board.
5
Measure vertical and horizontal
distances using a ruler.
6
Repeat up to 10 times and record
the results of the data in the data
tab for the results of the interim
report.
WORK STEPS (STATIC SHEET STYLE)
RUDE SURFACE.
NO PICTURE
OF NAME OF EQUIPMENT
AND MATERIALS
1
Prepare a tool for practicing
static and kinetic friction forces.

9. 2
Weigh the mass of the rough
surface.
NO PICTURE
OF NAME OF EQUIPMENT
AND MATERIALS
3
Prepare a tool for practicing
static and kinetic friction forces.
NO IMAGES
OF NAMES TOOLS AND
MATERIALS
4
Lift the inclination board with the
inclination load brace and place
the rough beams on the
inclination board.
5
Measure vertical and horizontal
distances using a ruler.

10. 6
Repeat up to 10 times and record
the results of the data in the data
tab for the results of the interim
report.
WORK STEPS (KINETIS STYLE STYLE)
LICIN SURFACE.
NO PICTURE
OF NAME OF EQUIPMENT
AND MATERIALS
1
Prepare a tool for practicing
static and kinetic friction forces.
2
Prepare the inclination board
thanks to

11. 3
Measure board length inclination.
50 cm.
4
Prepare a stopwatch to measure
time.
5
Place the slippery beams on the
inclination board, and the heavier
beams are hung in the pulley
then release them together with
the stopwatch.
NO IMAGES
OF NAMES TOOLS AND
MATERIALS
6
Repeat up to 10 times and record
the results of the data in the data
tab for the results of the interim
report.
WORK STEPS (KINETIS STYLE STYLE)
RUDE SURFACE.
NO PICTURE
OF NAME OF EQUIPMENT
AND MATERIALS

12. 1
Prepare a tool for practicing
static and kinetic friction forces.
2
Prepare the inclination board
thanks to
3
Measure board length inclination.
50 cm.
4
Prepare a stopwatch to measure
time.
NO PICTURES
OF NAMES TOOLS AND
MATERIALS
5
Place the rough surface beams on
the inclination board, and the
heavier beams are hung in the
pulley then release them
together with the stopwatch.

13. 6
Repeat up to 10 times and record
the results of the data in the data
tab for the results of the interim
report.
E. DATA EXPERIMENT
Experiment I (Static Swipe Coefficient Measurement).
a. Slippery Surface.
Deuteronomy x (cm) y (cm) 𝜇s
1 18.8 10 0.5319
2 17.5 10 0.5714
3 16.1 10 0.6211
4 17.5 10 0.5714
5 17.9 10 0.5586
6 15 10 0.6666
7 17.6 10 0.5681
8 16 10 0.625
9 16.8 10 0.5952
10 19.3 10 0.5181
Rearata 17.25 10 0.5827
b. Rough surface
Deuteronomy x (cm) y (cm) 𝜇s
1 12.5 10 0.8
2 10.5 10 0.9523
3 12.2 10 0.8196

14. 4 10.9 10 0.9174
5 12 10 0.8333
6 9 10 1.1111
7 10.6 10 0.943
8 12.1 10 0.8264
9 12.8 10 0.7812
10 10.3 10 0.9708
Rearata 11.29 10 0.8955
Experiment II (Measurement of kinetic swipe coefficient).
a. Surface
⮚ Table I
Deuterono
my
Mileage x (cm) Time t (s)
Perceptan (m /
s2)
1 50 0.78 1.6436
2 50 0.53 3.5599
3 50 0.65 2.3668
4 50 0 , 59 2.8727
5 50 0.81 1.5241
6 50 0.82 1.4872
7 50 0.75 1.7777
8 50 0.82 1.4872
9 50 0.84 1.4172
10 50 0.75 1.7777
Rearata 50 0.734 1.99141
⮚ Table II
Kinetic coefficient Swipe
mean
1.9914
SD
1.9914 ± 0.0714

15. 2 ± 0.8
⮚ Table III
Rope Voltage
Average
0.9369 N
SD
0.9369 ± 0.0519
1 ± 0.05
b. Rough Surface
⮚ Table I
Deuteronomy
Mileage x
(cm)
Time t (cm)
Perceptan (m /
s2)
1 50 0.75 1.7777
2 50 0.63 2.5195
3 50 0.94 1.1317
4 50 0.63 2.5195
5 50 0.88 1.2913
6 50 0.78 1.6436
7 50 0.94 1.1317
8 50 0.75 1.7777
9 50 0, 68 2.1626
10 50 0.69 2,1004

16. Rearata 50 0.767 1.8056
⮚ Table II
Kinetic Swipe Coefficient
Average
1.3217.
SD
1.3217± 0.0519
1.3 ± 0.05
⮚ Table III
Rope Voltage
Average
0.9592 N
SD
0.9592± 0.0197
1.0 ± 0.02
F. MANAGEMENT DATA
Experiment I (Static Swipe Coefficient Measurement).

17. Slippery surface.
1. Dik:
𝑥 = 18.8 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
18.8
𝜇 𝑠 = 0.5319
2. Dik:
𝑥 = 17.5 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
17.5
𝜇 𝑠 = 0.5714
3. Dik:
𝑥 = 16.1 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
16,1
𝜇 𝑠 = 0.6211
4. Dik:
𝑥 = 17.5 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
17.5
𝜇 𝑠 = 0.5714
5. Dik:
𝑥 = 17.9 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
17.9
𝜇 𝑠 = 0.5587
6. Dik:
𝑥 = 15 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
15
𝜇 𝑠 = 0.6667
7. Dik:
Rough surface.
1. Dik:
𝑥 = 12.5 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
12.5
𝜇 𝑠 = 0.8
2. Dik:
𝑥 = 10.5 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
10,5
𝜇 𝑠 = 0.9523
3. Dik:
𝑥 = 12.2 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
12.2
𝜇 𝑠 = 0.8196
4. Dik:
𝑥 = 10.9 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
10.9
𝜇 𝑠 = 0.9174
5. Dik:
𝑥 = 12 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
12
𝜇 𝑠 = 0.8333
6. Dik:
𝑥 = 9 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
9
𝜇 𝑠 = 1,1111
7. Dik:

18. 𝑥 = 17.6 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
17.6
𝜇 𝑠 = 0.5682
8. Dik:
𝑥 = 16 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
16
𝜇 𝑠 = 0.625
9. Dik:
𝑥 = 16.8 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
16.8
𝜇 𝑠 = 0.5952
10. Dik:
𝑥 = 19.3 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
19.3
𝜇 𝑠 = 0.5181
𝑥 = 10.6 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
10.6
𝜇 𝑠 = 0.9434
8. Dik:
𝑥 = 12.1 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
12,1
𝜇 𝑠 = 0.8264
9. Dik:
𝑥 = 12.8 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
12.8
𝜇 𝑠 = 0.7812
10. Dik:
𝑥 = 10.3 𝑐𝑚
𝑦 = 10 𝑐𝑚
Dit: 𝜇s
𝜇 𝑠 =
𝑦
𝑥
𝜇 𝑠 =
10
10.3
𝜇 𝑠 = 0.9708
Determine the maximum static friction force.
The formula fs = 𝜇 𝑠. 𝑁 𝑤𝑖𝑡ℎ 𝑁 = 𝑤 = 𝑚𝑔
Slippery surface.
1. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.5319
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.5319 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.3388 𝑁
2. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.5714
Surface.
1. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.8
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.8 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.5096 𝑁
2. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.9523

19. Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.5714 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.3639 𝑁
3. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.6211
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.6211 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.3956 𝑁
4. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.6211
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.5714 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.3639 𝑁
5. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.5587
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.5587 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.3558 𝑁
6. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.6667
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.6667 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.4246 𝑁
7. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.5682
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.5682 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.3619 𝑁
8. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.625
Dit: fs
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.9523 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.6066 𝑁
3. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.8196
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.8196 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.5220 𝑁
4. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.9174
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.9174 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.5843 𝑁
5. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.8333
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.8333 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.5308 𝑁
6. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 1.1111
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 1.1111 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.7077 𝑁
7. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.9434
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.9434 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0,6009 𝑁
8. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.8264
Dit: fs

20. fs= 𝜇 𝑠 𝑚𝑔
fs= 0.625 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.3981 𝑁
9. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.5952
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.5952 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.3791 𝑁
10. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.5181
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.5181 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.3300 𝑁 𝑅𝑜𝑢𝑔ℎ
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.8264 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.5264 𝑁
9. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.7812
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.7812 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.4976 𝑁
10. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑠 = 0.9708
Dit: fs
fs= 𝜇 𝑠 𝑚𝑔
fs= 0.9708 𝑥 65 𝑥 10−3
𝑥 9.8
fs= 0.6184 𝑁
Experiment II (Measurement of Kinetic Swipe Coefficient).
Calculateacceleration
slippery surface.
1. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.78 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.6084
𝑎 = 1.6436 𝑚. 𝑠−2
2. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.53 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.2809
𝑎 = 3.5599 𝑚. 𝑠−2
3. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.65 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.4225
𝑎 = 2.3668 𝑚. 𝑠−2
4. Dik:
Rough surface.
1. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.75 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.5625
𝑎 = 1.7777 𝑚. 𝑠−2
2. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.63 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.3969
𝑎 = 2.5195 𝑚. 𝑠−2
3. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.94 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.8836
𝑎 = 1.1317 𝑚. 𝑠−2
4. Dik:

21. 𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.59 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.3481
𝑎 = 2.8727 𝑚. 𝑠−2
5. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.81 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.6561
𝑎 = 1.5241𝑚. 𝑠−2
6. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.82 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.6724
𝑎 = 1.4872𝑚. 𝑠−2
7. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.75 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.5625
𝑎 = 1.7777𝑚. 𝑠−2
8. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.82 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.6724
𝑎 = 1.4872𝑚. 𝑠−2
9. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.84 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.7056
𝑎 = 1.4172𝑚. 𝑠−2
10. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.75 𝑠
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.63 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.3969
𝑎 = 2.5195 𝑚. 𝑠−2
5. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.88 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.7744
𝑎 = 1.2913 𝑚. 𝑠−2
6. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.78 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.6084
𝑎 = 1.6436 𝑚. 𝑠−2
7. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.94 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.8836
𝑎 = 1.1317 𝑚. 𝑠−2
8. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.75 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.5625
𝑎 = 1.7777 𝑚. 𝑠−2
9. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.68 𝑠
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.4124
𝑎 = 2.1626 𝑚. 𝑠−2
10. Dik:
𝑥 = 50 𝑐𝑚 = 0.5 𝑚
𝑡 = 0.69 𝑠

22. Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.5625
𝑎 = 1.7777𝑚. 𝑠−2
Dit:
𝑎 =
2𝑠
𝑡2
𝑎 =
2 𝑥 0.5
0.4761
𝑎 = 2,1004 𝑚. 𝑠−2
Calculates kinetic friction coefficient
Slippery surface.
1. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.6436 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.6436
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1677(2.8461)
𝜇 𝑘 = 1.3688
2. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 3.5599 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
3.5599
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.3632(2.8461)
𝜇 𝑘 = 0.8122
3. Dik:
Rough surface.
1. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.7777 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.7777
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1813(2.8461)
𝜇 𝑘 = 1.3298
2. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 2.5195 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
2.5195
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.2570(2.8461)
𝜇 𝑘 = 1.1144
3. Dik:

23. 𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 2.3668 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
2.3668
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.2415(2.8461)
𝜇 𝑘 = 1.1587
4. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 2.8727 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
2.8727
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.2931(2.8461)
𝜇 𝑘 = 1.0118
5. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.5241 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.5241
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1555(2,8461)
𝜇 𝑘 = 1.4035
6. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.1317 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.1317
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1154(2.8461)
𝜇 𝑘 = 1.5174
4. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 2.5195 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
2.5195
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.2570(2.8461)
𝜇 𝑘 = 1.1144
5. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.2913 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.2913
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1317(2.8461)
𝜇 𝑘 = 1.4711
6. Dik:

24. 𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.4872 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.4872
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1517(2.8461)
𝜇 𝑘 = 1.4142
7. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.7777 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.7777
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1813(2.8461)
𝜇 𝑘 = 1.3298
8. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.4872 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.4872
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1517(2.8461)
𝜇 𝑘 = 1.4142
9. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.6436 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.6436
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1677(2.8461)
𝜇 𝑘 = 1.3688
7. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.1317 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.1317
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1154(2.8461)
𝜇 𝑘 = 1.5174
8. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.7777 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.7777
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1813(2.8461)
𝜇 𝑘 = 1.3298
9. Dik:

25. 𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.4172 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.4172
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1446(2.8461)
𝜇 𝑘 = 1.4345
10. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 1.7777 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
1.7777
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.1813(2.8461)
𝜇 𝑘 = 1.3298
Average Kinetis friction coefficient
on slippery plane.
𝜇 𝑘1 + 𝜇 𝑘2 + 𝜇 𝑘3 + 𝜇 𝑘4 + 𝜇 𝑘5 + 𝜇 𝑘6 +
𝜇 𝑘7 + 𝜇 𝐾8 + 𝜇 𝐾9 + 𝜇 𝐾10 =.
1.6436 + 3.5599 + 2.3668 +
2.8727 + 1.5241 + 1.4872 +
1.7777 + 1.4872 + 1.4172 +
1.7777 = 1.9914
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 2.1626 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
2.1626
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.2206(2.8461)
𝜇 𝑘 = 1.2180
10. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑎 = 2,1004 𝑚. 𝑠−2
𝜇 𝑘 =
𝑚2
𝑚1
−
𝑎
𝑔
(𝑚1+𝑚2)
𝑚1
𝜇 𝑘
=
0.12
0.065
−
2.1004
9.8
(0.065 + 0.12)
0.065
𝜇 𝑘 = 1.8461 − 0.2143(2.8461)
𝜇 𝑘 = 1.2361
The mean coefficient of friction
Kinetics in rough fields.
𝜇 𝑘1 + 𝜇 𝑘2 + 𝜇 𝑘3 + 𝜇 𝑘4 + 𝜇 𝑘5 + 𝜇 𝑘6 +
𝜇 𝑘7 + 𝜇 𝐾8 + 𝜇 𝐾9 + 𝜇 𝐾10 =.
1.3298 + 1.1144 + 1.5174 +
1.1144 + 1.4711 + 1.3688 +
1.5174 + 1.3298 + 1.2180 +
1.2361 = 1.3217.
Calculates the standard deviation of the kinetic friction coefficient on slippery
fields.
𝑥𝑖 𝑥 (𝑥 − 𝑥𝑖) (𝑥 − 𝑥𝑖)2
1.6436 1.9914 0.3478 0.12096
3.5599 1.9914 -1.5685 2.46019
2.3668 1.9914 -0.3754 0.14093
2.8727 1.9914 -0.8813 0, 77669

26. 1.5241 1.9914 0.4673 0.21837
1.4872 1.9914 0.5042 0.25422
1.7777 1.9914 0.2137 0.04567
1.4872 1.9914 0.5042 0.25422
1 ,
4172
1.9914 0.5742 0.32971
1.7777 1.9914 0.2137 0.04567
√∑10
𝑖 = 1 𝑥−𝑥 𝑖
2
10(10−1)
=
0.4646
90
= √0.0051
SD = 0.0714
0.46466
Calculates the standard deviation of the kinetic friction coefficient in the rough
field of
Data 𝑥𝑖 𝑥
(𝑥 −
𝑥𝑖)
(𝑥 − 𝑥𝑖)2
1 1.7777 1.8055 0.0278 0.00077
2 2.5195 1.8055 -0.714 0.5098
3 1.1317 1.8055 0.6738 0.45401
4 2.5195 1.8055 -0.714 0.5098
5 1.2913 1.8055 0.5142 0.2644
6 1.6436 1.8055 0.1619 0.02621
7 1.1317 1.8055 0.6738 0.45401
8 1.7777 1.8055 0.0278 0 , 00077
9 2,1626 1,8055 -0.3571 0.12752
10 2,1004 1,8055 -0.2949 0.08697
√∑10
𝑖 = 1 𝑥−𝑥 𝑖
2
10(10−1)
=
0.2434
90
= √0.0027
SD = 0.0519
0.2434

27. Counting Rope Voltage kinetic coefficient of friction on a rough field
with the formula equation: 𝐸 =
((𝑚1+𝑚2)𝑔𝜇(𝑘+ 1)
(𝑚1+𝑚2)
surface is slippery.
1. Dik:
𝑚1 = 65 𝑔𝑟𝑎𝑚𝑠 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.3688
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.3688 + 1)
(0.065 + 0.12)
𝑇 = 0.9787 𝑁
2. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 0.8122
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (0.8122 + 1)
(0.065 + 0.12)
Rough surface.
1. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.3298
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.3298 + 1)
(0.065 + 0.12)
𝑇 = 0.9626 𝑁
2. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.1144
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.1144 + 1)
(0.065 + 0.12)

28. 𝑇 = 0.7487 𝑁
3. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.1587
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.1587 + 1)
(0.065 + 0.12)
𝑇 = 0.8919 𝑁
4. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.0118
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.0118 + 1)
(0.065 + 0.12)
𝑇 = 0.8312 𝑁
5. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.4035
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.4035 + 1)
(0.065 + 0.12)
𝑇 = 0.9931 𝑁
6. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝑇 = 0.8736 𝑁
3. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.5174
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.5174 + 1)
(0.065 + 0.12)
𝑇 = 1.0401 𝑁
4. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.1144
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.1144 + 1)
(0.065 + 0.12)
𝑇 = 0.8736 𝑁
5. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.4711
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.4711 + 1)
(0.065 + 0.12)
𝑇 = 1.0210 𝑁
6. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2

29. 𝜇 𝑘 = 1.4142
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.4142 + 1)
(0.065 + 0.12)
𝑇 = 0.9975 𝑁
7. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.3298
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.3298 + 1)
(0.065 + 0.12)
𝑇 = 0.9626 𝑁
8. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.4142
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.4142 + 1)
(0.065 + 0.12)
𝑇 = 0.9975 𝑁
9. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.4345
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.4345 + 1)
(0.065 + 0.12)
𝑇 = 1.0059 𝑁
10. Dik:
𝜇 𝑘 = 1.3688
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.3688 + 1)
(0.065 + 0.12)
𝑇 = 0.9787 𝑁
7. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.5174
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.5174 + 1)
(0.065 + 0.12)
𝑇 = 1.0401 𝑁
8. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.3298
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.3298 + 1)
(0.065 + 0.12)
𝑇 = 0.9626 𝑁
9. Dik:
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.2180
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.2180 + 1)
(0.065 + 0.12)
𝑇 = 0.9164 𝑁
10. Dik:

30. 𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.3298
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.3298 + 1)
(0.065 + 0.12)
𝑇 = 0.9626 𝑁
𝑚1 = 65 𝑔𝑟 =
65 𝑥 10−3
𝑘𝑔
𝑚2 = 120 𝑔𝑟 =
120𝑥10−3
𝑘𝑔
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.2361
Dit: rope tension.
𝑇 =
((𝑚1+𝑚2)𝑔 (𝜇 𝑘+1)
(𝑚1+𝑚2)
𝑇 =
((0.065 + 0.12)9.8 (1.2361 + 1)
(0.065 + 0.12)
𝑇 = 0.9239 𝑁
Average Kinetis friction coefficient
on slippery plane.
𝑇1 + 𝑇2 + 𝑇3 + 𝑇4 + 𝑇5 + 𝑇6 + 𝑇7 +
𝑇8 + 𝑇9 + 𝑇10 =.
0.9787 + 0.7487 + 0.8919 +
0.8312 + 0.9931 + 0.9975 +
0.9626 + 0.9975 + 1.0059 +
0.9626 = 0.9369 N
The mean coefficient of friction
Kinetis in rough fields.
. 𝑇1 + 𝑇2 + 𝑇3 + 𝑇4 + 𝑇5 + 𝑇6 + 𝑇7 +
𝑇8 + 𝑇9 + 𝑇10 =.
0,9626 + 0,8736 + 1,0401 +
0,8736 + 1,021 + 0,9787 +
1,0401 + 0,9626 + 0,9164 +
0,9239 = 0,9592 N
Calculate the Standard of Voltage Devisionation Rope kinetic friction coefficient
in rough fields.
Data 𝑥𝑖 𝑥 (𝑥 − 𝑥𝑖) (𝑥 − 𝑥𝑖)2
1 0.9626 0.9592 -0.0034 0.00001156
2 0.8736 0.9592 0.0856 0.00732740
3 1.0401 0.9592 -0.0809 0.00654480
4 0.8736 0.9592 0,
0856
0.0073
2740 5
1.021
0.9592
-
0.0618
0.0038
1920 6

31. , 9592 -0.0034 0.00001156
9 0.9164 0.9592 0.0428 0.00183180
10 0.9239 0.9592 0.0353 0.00124610
√∑10
𝑖 = 1 𝑥−𝑥 𝑖
2
10(10−1)
=
0.03504
90
= √0,000389
SD = 0.0197
0.03504
Calculating the Standard of Voltage Devisionation Rope kinetic friction coefficient
on the slippery plane.
Data 𝑥𝑖 𝑥 (𝑥 − 𝑥𝑖) (𝑥 − 𝑥𝑖)2
1 0.9787
0.9369 -
0.0418
0.0017 2
0.7487
0.9369
,
0111 5
0.9931
0.9369
-
0.0562
0.0031
6
0.9975

32. -0.0606 0.0036
9 1.0059 0.9369 -0.069 0.0047
10 0.9626 0.9369 -0.0257 0.0006
√∑10
𝑖 = 1 𝑥−𝑥 𝑖
2
10(10−1)
=
0.2434
90
√0.0027
SD
=0.0519
0.0664
Determine the maximum kinetic friction force.
The formula fk = 𝜇 𝑘. 𝑁 𝑤𝑖𝑡ℎ 𝑁 = 𝑤 = 𝑚𝑔
Slippery surface.
1. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.3688
Dit: fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1.3668 𝑥 65 𝑥 10−3
𝑥 9.8
Fk= 0.8719 𝑁
2. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 0.8122
Dit: fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 0.8122 𝑥 65 𝑥 10−3
𝑥 9.8
Fk= 0.5173 𝑁
PermukaanKasar.
1. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,3298
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,3298 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,8470 𝑁
2. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,1144
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,1144 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,7098 𝑁

33. 3. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.1587
Dit: fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1.1587 𝑥 65 𝑥 10−3
𝑥 9.8
Fk= 0.7380 𝑁
4. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.0118
Dit: fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1.0118 𝑥 65 𝑥 10−3
𝑥 9.8
Fk= 0.6445 𝑁
5. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.4035
Dit: fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1.4035 𝑥 65 𝑥 10−3
𝑥 9.8
Fk= 0.8940 𝑁
6. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.4142
Dit: fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1.4142 𝑥 65 𝑥 10−3
𝑥 9.8
Fk= 0.9008 𝑁
7. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.3298
Dit: fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1.3298 𝑥 65 𝑥 10−3
𝑥 9.8
Fk= 0.8470 𝑁
8. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1.4142
Dit: fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1.4142 𝑥 65 𝑥 10−3
𝑥 9.8
Fk= 0.9008 𝑁
9. Dik:
3. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,5174
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,5174 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,9665 𝑁
4. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,1144
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,1144 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,7098 𝑁
5. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,4711
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,4711 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,9370 𝑁
6. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,4711
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,4711 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,9370 𝑁
7. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,5174
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,5174 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,9665 𝑁
8. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,3298
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,3298 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,8470 𝑁
9. Dik:

34. 𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9.8 𝑚 /𝑠2
𝜇 𝑘 = 1,4345
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,4345 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,9137 𝑁
10. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,3298
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,3298 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,8470 𝑁
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,218
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,218 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,7758 𝑁
10. Dik:
𝑚 = 65 𝑔 = 65 𝑥 10−3
𝑔 = 9,8 𝑚/𝑠2
𝜇 𝑘 = 1,2361
Dit : fk
Fk= 𝜇 𝑘 𝑚𝑔
Fk= 1,2361 𝑥 65 𝑥 10−3
𝑥 9,8
Fk= 0,7873 𝑁