Name: Vinh Vu Date: Nov. 19, 2015
Friction and Acceleration Write out a definition for each term below. Newton's First Law of Motion- If an object is at rest, it stays at rest unless acted upon
by a force. If an object is in motion, it moves in a constant velocity unless acted upon by
a force. Newton's Second Law of Motion- The acceleration of a moving object is produced by
a constant force. It is directly proportional to the net force and inversely proportional to
the mass of the object. The direction of acceleration is the same as the net force’s
direction. Newton's Third Law of Motion- For every action, there is a reaction force that is
equal in magnitude but opposite in direction. The action-reaction force always act on
two different objects. Friction – Friction is a force resisting the motion of an object that appears being an
object and the surface. Static Friction – It is a type of friction that keeps objects from moving or sliding. The
coefficient of static friction is higher than kinetic friction. Kinetic Friction – It is a type of friction that occurs when two objects are moving
relative to each other. They make contact when they are in motion. The coefficient of
kinetic friction is less than the coefficient of static friction.
Lab Activity – Online Simulation
Open up the University of Colorado, PhET "Forces and Motion" simulation
http://phet.colorado.edu/en/simulation/forces-and-motion-basics
Spend a few minutes familiarizing yourself with the controls of the simulation before you begin
the lab below.
Part I (Friction):
Click on the Friction Tab and check all the boxes in the upper right hand corner (Forces, Sum of
Forces, Values, Masses, Speed)
a. place 40 kg girl on the screen. Type in the box for the applied force 150 N. The screen should
look like: M. Mikhaiel September 16, 2013 b. Read the friction force and then calculate the weight of the girl.
Fg=mg
m=40kg
Fg= (40)9.8
Fg= 392N
c. Repeat (a) and (b) for other objects and construct a table like the one shown below:
Mass of Object (kg) Applied Force (N) 40
50
80
100
150 150
200
250
300
400 Friction (N) Weight of Object = Normal
Force 100
125
200
250
375 400
500
800
1000
1500 d. Plot a graph of Friction vs. Normal Force and find the slope of the graph. Friction vs Normal Force
400
350
300
250 Friction (N) 200
150
100
50
0
400 500 800 1000 1500 Normal Force (N)
Column2 e. What does the slope of the graph represent? The slope of the graph
represents the coefficient of friction. Since the formula for friction is
Ffric= µFN, the slope , rise/run, is Ffric/FN which stands for the coefficient
of friction, µ. f. what is the unit of the slope, if any? There is no unit to the slope because Ffric/FN is
Newton/Newton which cancels each other out. The unit makes sense because normally the
coefficient of friction does not have a unit. g. Challenge Question:
What is the mass of the gift box? Explain how you got your answer.
From the previous data points, the coefficient of the surface is 0.25 because Ffric/FN for all the
objects gives 0.25 for Mu. For the gift box, the kinetic frictional force is 125 N with a 150 N
applied force.
Ffric= µFN
FN= Fg
Ffric/µ= FN
FN=mg
FN= 500 N
FN/g=m
m= 50kg The mass of the gift box is 50kg as seen on the math. Part II (Acceleration):
Click on the Acceleration Tab and check all the boxes in the upper right hand corner (Forces,
Sum of Forces, Values, Masses, Speed, and acceleration). Don't change the friction scale shown
on your screen
a. Place 50 kg box on the screen. The screen should look like: Type in the box for the applied force 150 N and calculate the time it will take for the box to go to
the maximum speed. The maximum speed is reached when the hand on the speedometer cannot
go any further.
Open a new browser (for the purpose of measuring the time it will take for the box to go to the
maximum speed) and click on the link below for the stop watch.
http://www.online-stopwatch.com/full-screen-stopwatch/
You will see something similar like: b. Find the net force from the animation and calculate the acceleration of the box by using the
equation a = v/t. Use an average speed of 40 m/s
t= 85s
a= v/t
a= 40/85
a= 0.47 m/s2
c. Repeat (a) and (b) for applied forces of 200 N, 250 N, and 300 N.
200 N:
t= 30s
a= v/t
a= 40/30
a= 1.33 m/s2 250 N:
t= 18s
a= v/t
a= 40/18
a= 2.22 m/s2 300 N:
t= 12s
a= v/t
a= 40/12
a= 3.33 m/s2 d. Record your answers as shown in the table below:
Applied Force
(N)
150
200
250
300 Net Force (N)
25
75
125
175 Mass (kg)
50
50
50
50 Time (sec)
85
30
18
12 Average Speed
(m/s)
40
40
40
40 Acceleration
2
(m/s )
0.47
1.33
2.22
3.33 2 e. Plot a graph of Net Force (N) vs. Acceleration (m/s ). Net Force vs Acceleration
200
180
160
140
120 Net Force (N) 100
80
60
40
20
0
0.47 1.33 2.2200000000000002 3.33 Acceleration (m/s2)
Column3 f. Find the slope of the graph. What does the slope represent?
The slope of the graph represents the mass of the object. The slope of the graph is Net force/
acceleration; based on Newton’s Second Law, ƩF= ma and ƩF/a=m so the slope gives mass of
the object.
g. Find the % error of the mass of the object.
% of error= | (mexperiment – mtheoretical)/ mtheoretical | x 100
% of error= | (52.55-50)/50| x 100
% of error= 5% Conclusion:
Write a short conclusion of what you have learned from this experiment.
This lab has mapped out the connections between various variables that relate to Newton’s
Laws. It has helped me to have a further understanding of graphing force and determining
the variables for the axes. From the graphing activities, I have learned to utilize the unit of
each variable to determine the axes of the graphs. The lab has strengthened my knowledge
about force and its relationships.

 

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