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PUM Physics II - Dynamics
Lesson 9 Solutions
Page 1 of 4
9.1 Represent and Reason
Description
of the
object of
interest is
underlined
1) A 2.2 kg
bucket of
clams sits at
rest on a
desk.
A
Sketch the
situation.
Circle the
object of
interest.
Draw the
direction of the
acceleration, if
known.
Δv = 0
B
Translate the
givens into
physical
quantities.
C
Draw a force
diagram for the
object of interest.
D
Can you
evaluate
any of the
forces in the
force
diagram?
E
Write Newton’s Second Law in
component form.
Fill in anything you know and
solve for anything you do not
know.
Which are
negative
and which
are
positive?
Given in
description:
a = 0 (sits )
m = 2.2kg
FE on B y = mg=-22 N
ay 
0
FEarth on Bucket y  FTable on Bucket y
m
(22N)  FTable on Bucket
2.2kg
FTonB y = 22 N
2) A 5kg
bucket of
clams hangs
motionless
from a
spring that
stretches 40
cm.
3) A man
pulls a 40kg
refrigerator
up an
elevator
shaft with a
rope at a
constant
speed.
Come up
with your
own for an
object in
equilibrium
with 3 or
more other
objects
interacting
with it.
Given in
description:
a=0
(motionless)
m =5 kg
x = 0.4 m
FE on B y = mg= -(5
kg)(10 m/s2)
= -50 N
FSonB
FEonB
ay = (FE on B y + FSonB y)/m
0 m/s2 = [(-50 N) + FSonB y]/(5 kg)
FSonB y = kx = 50 N
k(0.4 m) = 50 N
k = (50 N)/(0.4 m) = 125 N/m
Δv = 0
Δv = 0
Given in
description:
a = 0 (const
velocity)
m =40 kg
FE on R y = mg= -(40
kg)(10 m/s2)
= -400 N
FRonR
ay = (FE on R y + FRonR y)/m
0 m/s2 = [(-400 N) + FRonR y]/(40
kg)
FRonR = 400 N
FEonR
Cat in spring
bucket:
mB = 5 kg
mc = 15 kg
k = 150 N/m
FSonB
Bucket is the
system
FEonB
FConB
FConB = -mg
= -(15
kg)(10 m/s2)
= -150 N
FEonB = -50
N
ay = (FSonB + FEonB + FConB)/mB
0 m/s2 = [FSonB + (-50 N) + (-150
N)]/(5kg)
FSonB = 200 N = kx
kx = (150 N/m)x = 200 N
x = 1.3 m
PUM Physics II - Dynamics
Lesson 9 Solutions
Page 2 of 4
9.2 Regular Problem
a)
Motion Diagram: Right is positive.
Force Diagram:
FFonC
Δv
FYonC
FEonC
b) Newton’s 2nd Law
ax = -FYonC/mC
c) ax = -FYonC/mC = (-12 N)/(14.5 kg) = -0.83 m/s2
vo = 2 m/s
x(t) = xo + vt + ½at2
x = (2 m/s)(3 s) + ½(-0.83 m/s2)(3 s)2 = 2.3 m
9.3 Regular Problem
Assuming there are no other forces exerted on the beam except by the astronaut.
ax = FAonB/mB
mB = FAonB/ax = (150 N)/(0.15 m/s2) = 1000 kg
9.4 Design an Experiment
a) Many Possible Experiments. For example, students could decide to pull with the same
force on various air gliders of different masses with spring scales to determine if the
acceleration depends on 1/m.
b) The best experiment should create the best chance of disproving the relationship a =
ΣF/m
c) Equipment and data will depend on the chosen experiment.
d) Students should use their experiment and Newton’s 2nd law to base their prediction. The
prediction should be quantitative.
e) Newton’s 2nd law is the hypothesis mathematically it states: a = ΣF/m . The prediction
should be qualitative or quantitative based on the experiment if Newton’s 2nd law is true.
Using the equation students can make a quantitative prediction based on their experiment,
putting in numbers for F and m to predict an acceleration for example.
PUM Physics II - Dynamics
Lesson 9 Solutions
Page 3 of 4
9.5 Reason
When the force of the cable on the elevator balances with the downward force of Earth on the
elevator, the sum of the forces that the cable and Earth exert on the elevator is now zero. The
acceleration then will also be zero. Therefore the elevator will continue moving at constant
velocity. This is assuming there is no friction or air resistance.
Forces balance
Forces balance
PUM Physics II - Dynamics
Lesson 9 Solutions
Page 4 of 4
9.6 Practice
Description
of the object
of interest is
underlined
A
Sketch the
situation.
Circle the
object of
interest.
Draw a
motion
diagram
and the
direction of
the
acceleration,
if known
B
Translate
the givens
into
physical
quantities.
1) A 72 kg
crate on a
freight
elevator
accelerates
upwards at a
rate of 0.2
m/s2 while
moving
down.
m = 72 kg
a = +0.2
m/s2
2) A 172 kg
crate on a
freight
elevator
accelerates
downwards
at a rate of
0.4 m/s2
while
moving up.
m = 172 kg
a = -0.2
m/s2
3) A physics
teacher of
mass m is
holding onto
a rope
attached to a
hot air
balloon and
is
accelerating
upwards at a
m/s2.
m
a
C
Draw a
force
diagram
for the
object of
interest.
FFEonC
D
Can you
evaluate any of
the force
components in
the force
diagram?
Which are
negative and
which are
positive? What
if you changed
the direction of
the axes?
FEonC = -mg = (72 kg)(10 m/s2)
= -720 N
Would be
positive if axis
was reversed.
FEonC
FEonC = -mg = (172 kg)(10 m/s2)
= -1720 N
FFEonC
E
Write Newton’s Second Law in
component form.
What can you determine using
the information in the problem?
ay = (FFEonC + FEonC)/m
0.2 m/s2 = [FFEonC + (-720 N)]/(72
kg)
FFEonC = (0.2 m/s2)(72 kg) + 720 N
= 734 N
ay = (FFEonC + FEonC)/m
-0.4 m/s2 = [FFEonC + (-1720
N)]/(172 kg)
Would be
positive if axis
was reversed.
FFEonC = -(0.4 m/s2)(172 kg) +
1720 N = 1651 N
FEonPT = -mg
a = (FBonPT + FEonPT)/m
Would be
positive if axis
was reversed.
FBonPT = ma - FEonPT
FEonC
FBonPT
FBonPT = ma + mg
FBonPT = m(a + g)
FEonPT