Download Kinematics and Newton`s 2nd Law Key

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Key
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Net Force Particle Model Worksheet 3:
Kinematics & Newton's 2nd Law
The problems on the worksheet require you to use kinematics formulas in addition to Newton's
second law. Use the following steps in your solutions:
a. use force diagram analysis to find the net (unbalanced) amount of force.
b. list knowns and unknowns for force and motion variables:
force variables
motion variables
acceleration
mass
net force
acceleration
initial velocity
final velocity
change in time
displacement
mathematical model
mathematical models
vf = at + vi
Fnet
Δx=vt 
a m
x = ½at2 + vit
vf2 = vi2+ 2ax
c. The variable that ties both lists of variables together is acceleration. Depending on the
variables you know, use either the force or motion mathematical models to solve for
acceleration, and then use the acceleration value to solve for the unknown quantity.
1. A race car has a mass of 710 kg. It starts from rest and travels 40.0m in 3.0s. The car is
uniformly accelerated during the entire time. How big is the net force acting on the car?
Make a quantitative force diagram. Write a net force equation for the axis along which
forces are not balanced.
vf = ___ x = 40 m
vi = 0.0 m/s
data
time velocity
0.0 0.0
1.0 ____
2.0 ____
3.0 26.6
a = ___
Fnet = ____
position
0.0
___
____
_40_
x 40  0m
v 

 13 .3 m s
t
3s
v v
v  i f  v f  2v v i
2
v f  2 13 .3 m s  26 .6 m s

t = 3.0 s

FN
Ff,car,road
m
v 26 .6  0 ms

 8.9 s
t
3s
s
Fnet
a
 Fnet  ma
m


Froad  710 k g 8.9 m 2  6320 N

s 
a
Fg

©Modeling Instruction 2013

1
U5 Net Force – ws3 v3.1
2. Suppose that a 1000 kg car is traveling at 25 m/s (55 mph). Its brakes can apply a force of
5000 N. What is the minimum distance required for the car to stop? Make a quantitative
force diagram. Write a net force equation for the axis along which forces are not
balanced.
vi = 55 m/s
vf = 0.0 m/s
a = ___
Fbrakes = 5000 N
Fbrakes
a
Fnet
m
a
a
5000 N
1000 kg
a  5.0
FN
Ff
m
Fbrake
or  5.0
m
s2
m
s
Fg
each second
3. A 65 kg person dives into the water from the 10 m platform.
a. What is her speed as she enters the water?
vi = 0.0 m/s
vf = ____
y = -10 m
a = -10 m/s/s
v f2  v i2  2ay
v f2  0  2( 10 sm2 )( 10m)
v f2  200 ms 2  v f   200  14 ms
2
b. She comes to a stop 4.0 m below the surface of the water. Find the force on the swimmer
by the water.
vi = 14.14 m/s
vf = 0.0 m/s
y = - 4.0 m
a = ____
v f2  v i2  2ay
a
v f2 v i2
2 y
a
0 ( 14.14 ms )2
2( 4 m )
2

Fwater
0(200 m2 )
s
8 m
Fnet
m
a  25 sm2  25 s s
a
Fnet
m

Fwater Fg
m
 Fwater  Fg  ma
Fg
Fw  ma  Fg
N
Fwater  65kg (25 sm2 )  65kg (10 kg
)  1625N  650N  2275N  2300N
©Modeling Instruction 2013
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U5 Net Force – ws3 v3.1
4. During a head-on collision, a passenger in the front seat of a car accelerates from 13.3 m/s
(30 miles/hour) to rest in 0.10 s.
a. Calculate the acceleration of the passenger.
vi = 13.3 m/s
vf = 0.0 m/s
t = 0.10 s
a = ____
assume passenger in left car and to right is positive
a
v
t
a
m
0  13.3 ms
 133 sm2  133 ss
.10s
b. The driver of the car holds out his arm to keep his 25 kg child (who is not wearing a seat
belt) from smashing into the dashboard. How much force must he exert on the child?
vi = 13.3 m/s
a
Fnet
m
vf = 0.0 m/s

FN
m
t = 0.10 s
a = - 133 m/s/s (from part a.)
 FN  ma
m = 25 kg
Fnet
FN  25kg ( 133 )  3325N
m
s2
Farm
c. What is the weight of the child?
N
Fg  mg  25kg (10 kg
)  250N
d. Convert the forces in parts b and c from Newtons to pounds. (1 lb = 4.45N). What are the
chances the driver will be able to stop the child?
1lb
)  747 pounds
4.45N
1lb
250N(
)  56 pounds
4.45N
3325N(
The driver has no chance of stopping the child.
©Modeling Instruction 2013
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U5 Net Force – ws3 v3.1