Name
Date
Pd
Net Force Particle Model Worksheet 1:
Force Diagrams and Net Force
1. An elevator is moving up at a constant velocity of 2.5 m/s, as illustrated in the diagram
below: The passenger has a mass of 85 kg.
a. Construct a force diagram for the passenger.
y
F
=N
Fnet
=0
x
=
F
g
b. Calculate the force the floor exerts on the passenger.
FN = –Fg = –mg = -(85kg(-10 N/kg)) = 850 N
2.
The elevator now accelerates upward at 2.0 m/s2.
a. Construct a force diagram for the passenger.
F
ne
y
F
N
t
x
F
g
b. Write an equation for the vertical forces on the passenger.
Fnet = FN + Fg = ma
c. Calculate the force the floor exerts on the passenger.
Fg is -850N because it is directed downwards.
Fnet  FN  Fg  m a



FN  m a  Fg  85kg 2.0 sm2  850N  1020N
 ©Modeling Instruction 2013
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3.
Upon reaching the top of the building, the elevator accelerates downward at 3.0 m/s2.
a. Construct a force diagram for the passenger.
y
Fnet
=0
F
N
x
F
g
b. Write an equation for the vertical forces on the passenger.
Fnet = FN+ Fg = ma
c. Calculate the force the floor exerts on the passenger.
Fnet  FN  Fg  m a



FN  m a  Fg  85kg 3.0 sm2  850N  595N

4. While descending in the elevator, the cable suddenly breaks. How big is the force on the
passenger by the floor? Explain your answer.
The force on the passenger by the floor goes to 0 when the cable breaks because the elevator
is now in "free fall". Since the elevator and the passenger are falling together, the floor no
longer pushes up on the passenger. Algebraically, this is given by
Fnet  FN  Fg  m a


FN  m a  Fg  85kg 10.0 m 2  850N  0N

s 
5. a. A 70 kg skydiver jumps out of an airplane. Immediately after jumping, how large is the
skydiver's acceleration?


Immediately
after jumping, the only force on the skydiver is Fg which produces an

acceleration of 10 m/s/s
b. Upon reaching a downward velocity of 100 miles per hour, 300 newtons of drag resist the
diver's motion. Draw a force diagram for the skydiver. How large is the skydiver's
acceleration?
y
F
net
Fnet = Fdrag + Fg
Fdr
ag
Fnet = 300N - (70kg(10m/s/s))
Fnet = 300N - 700N = -400 N
F
m
F
400N
a  net 
 5.71 s
m
70kg
s

©Modeling Instruction 2013
x
g
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6. a. Draw a force diagram for a 900 kg car that exerts 5000 N of traction force on a level road
while being opposed by 1000 newtons of friction and drag forces
y
combined.
F
n
F
F
=
et
N
Ftracti g
x
F
on
=
f
F
g
b. Write a net force equation for the car.
Considering only forces in the x-direction
Fnet = Ftraction - F drag
c. Calculate the acceleration of the car.
a

m
Fnet (5000N 1000N)

 4.44 s
m
900kg
s
7. The three modified Atwood's machines shown below have blocks of mass M on a frictionless surface
and hanging from a string. When the blocks are released, they accelerate as they did in our lab.
a. Which system has the greatest net force? Explain how you know.
System C has the greatest net force. FN and Fg cancel out for the block moving
horizontally. All of the "FT"s cancel out inside the systems because each string is pulling
equally in both directions. The net force on each system is the force of gravity on the
hanging masses. System C has the heaviest hanging masses, so the net force acting on
the system is 2Mg.
b. Which system has the least inertia? Explain how you know.
If we define inertia as resistance to change in motion, and we know that in these systems
the resistance is determined by the mass of the system, then the system with the least
mass will have the least inertia. System A has the least mass.
c. Determine the acceleration for each system.
©Modeling Instruction 2013
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In each of these systems, the Fnet is provided by the weight, mg, of the hanging block(s).
System A, a 
m
Fnet Mg g

  5.0 s
m
2M 2
s
System B, a 
m
Fnet Mg g

  3.3 s
m
3M 3
s
m
F
2Mg 2g
System C, a  net 

 6.7 s
m
3M
3
s

8.
A child takes a trip down a slide.
 a quantitative force diagram for the 30kg child. The frictional force is 160 N.
a. Draw
Fnet
Ffk=1
F
FN=2
46N
g
Fg
║
Fg=30
b. Write an equation for the forces on the child parallel to the slide and find the net force on
the child.
Fnet = Fg parallel + Ffk


N
Fnet  mgsin55 160N  30kg 10 kg
 0.819 160N  86N down the slide
c. Calculate child's acceleration.

a
m
Fnet
86N

 2.87  2.9 s
m
30kg
s

©Modeling Instruction 2013
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