Examples and Hints in
Chapter 6
6.41

A small glider is placed against a compressed spring
at the bottom of an air track that slopes upward at an
angle of 400 above the horizontal. The glider has a
mass of 0.09 kg. The spring has a spring constant,
k=640 N/m and negligible mass. When the spring is
released, the glider travels a maximum distance of 1.8
m along the air track before sliding back down. Before
reaching this maximum distance, the glider loses
contact with the spring.
 A) What distance was the spring originally
compressed?
 B) When the glider has traveled along the air track 0.8
m from its initial position again the compressed spring,
is it still in contact with the spring? What is the kinetic
energy of the glider at this point?
Things to note
Air track– No friction
 Work done by spring is counteracted by
the work done by gravity.
 Work done by gravity can be calculated
by F*distance*cosine(angle between
them).

mg
L
900-400
400
W= -mg*L*cos(500)
W= -0.09*9.8*1.8*.647
W= -1.02 J
DK=Wtotal
In this case, DK=0 so Wtotal=0
 Wtotal=Wspring+ Wgravity
 Wtotal= ½ kX2 -1.02 J=0
 X2=1.02*2/640=0.031 m2
 X=0.056 m

Part B
Wtotal= ½ mv2-0=½ mv2=K
 Wgravity=-0.09*9.8*0.80*cos(500)=-0.45 J
 Wspring= ½ k X2 =0.5*640*.0562=1.003 J
 Wtotal= 1.003-0.45=0.5499 J=K
 No longer in contact with spring after 5.6
cm

6.69
A small block with a mass of 0.12 kg is attached to a cord
passing through a hole in a frictionless, horizontal
surface. The block is originally revolving at a
distance of 0.4 m from the hole with a speed of 0.70
m/s . The cord is then pulled from below, shortening
the radius of the circle in which the block revolves to
0.10 m. At this new distance, the speed of the block
is observed to be 2.8 m/s.
A) What is the tension in the cord in the original
situation when the block has a speed of 0.7 m/s?
B) What is the tension in the cord in the final situation
when the speed of the block is 2.8 m/s?
C) How much work was done by the person who pulled
on the cord?
Freebody Diagrams
2
mv
 F T
R
T
mv2/R
Part a)
If m=0.12 kg and v=0.7 m/s
and R=0.4 m then T must be
0.15 N
Part b)
If v=2.8 m/s and R=0.1 m then
T must be 9.41 N
Part C)

DK=W so
K2= ½ mv22 = 0.5 * 0.12kg*2.8^2=0.47 J
 K1= ½ mv12 = 0.5 * 0.12kg*0.7^2=0.0294 J
 Dk=K2-K1=0.47-0.294=0.441 J

Wild Monkey II
A monkey starts pushing a 400 N crate of bananas from
rest up an inclined plane whose surface is 300 above
the horizontal. The coefficient of kinetic friction of the
surface is 0.4. This is one smart monkey since he is
simultaneously lifting the box while pushing the box in
order to decrease friction. The monkey’s resultant
force makes an angle of 200 with the surface of the
inclined plane. The monkey’s force is NOT enough to
lift the box away from the surface so it stays in contact.
After the monkey has moved the box through a
distance of 5 m, the box’s velocity is 3 m/s. What is
the magnitude of the force that monkey is applying to
the box in newtons?
Draw It!
v2=3 m/s
200
F
v1=0
5m
f
300
Free body Diagrams
F*sin(200)
n
200
F*cos(200)
400 cos(300)
v2=3 m/s
200
F
f
v1=0
300
5m
f
400 sin(300)
300
Derive Equations
n
0=n+F*sin(200)-400*cos(300)
n= 400*cos(300)-F*sin(200)
F*cos(200)
400 cos(300)
f=mk*(400*cos(300)-F*sin(200))
f
400 sin(300)
Fnet=F*cos(200)+0.4*F*sin(200)-0.4*400*cos(300)-400*sin(300)
Fnet=1.07*F-338.5
Use Work Energy
n
Dk=K2-K1
K1=0
m=400/9.8=40.8 kg
K2= 0.5*40.8*3^2
K2= 183.6 J
Dk=183.6 J
Fnet=1.07*F-338.5
Wnet= Fnet * d
Wnet= (1.07*F-338.5)*5 m
Wnet=5.38*F-1692.8
F*sin(200)
F*cos(200)
400 cos(300)
f
400 sin(300)
Dk=Wnet
183.6 = 5.38*F-1692.8
1876.4=5.38*F
348.6N=F
Hints





6-22: Make a free body diagram of the forces, find the net work and use the work-energy
thm (W=DK)
6-26: Break weight into components parallel and perpendicular to the surface and then
use the work-energy thm.
6-36: Use the formula found on page 221 (between 6.9 and 6.10) and on the second part,
use the work energy thm
6-60: Part a) use only the work done by gravity to calculate the mass. Part b) use work
done by the normal force to calculate the force. Part c) W total=(m*a)*d where d=distance,
a=acceleration, m=mass and Wtotal=total work
6-67: Part a) Plug and chug to find the force at that point. Part b) Plug and chug again.
Part c) Use the anti-derivative or integral to integrate the force over x through the interval.
x2




1
6-70: Part a)  x
2
dx 
1 1

x1 x2
and then use the work-energy thm. Part b) Set up workenergy thm as before and let K2=0 Part c) The total work is zero.
6-72: Remember the scalar product is F*distance*cosine(angle between them)
6-80: See Hint on 6-72 and the students work is F*distance*cosine(angle between them).
The distance is parallel to the incline. Remember to include the component of weight
parallel to the surface.
6-82: Calculate the total work on both the horizontal block (work done by friction) and sum
that with the work done by gravity on the vertical block. W=0.5*(sum of
masses)*velocity^2
x1