6.1: a)
J60.3)m5.1()N40.2( 
b)
J900.0)m50.1)(N600.0( 
c)
J70.2J720.0J60.3 
.
6.2: a) “Pulling slowly” can be taken to mean that the bucket rises at constant speed, so
the tension in the rope may be taken to be the bucket’s weight. In pulling a given length
of rope, from Eq. (6.1),
J.6.264)m00.4)(s/m80.9()kg75.6(
2
 mgsFsW
b) Gravity is directed opposite to the direction of the bucket’s motion, so Eq. (6.2)
gives the negative of the result of part (a), or
J265
. c) The net work done on the bucket
is zero.
6.3:
J300)m0.12)(N0.25( 
.
6.4: a) The friction force to be overcome is
,N5.73)s/m80.9)(kg0.30)(25.0(
2
kk
 mgnf

or 74 N to two figures.

b) From Eq. (6.1),
J331)m5.4)(N5.73( Fs

extra figure. c) The normal force is

sinFmg 
, and so the work done by friction is
J5.386)30sin)N2.99()s/m80.9)(kg30)((25.0)(m50.4(
2

. d) Both the normal
force and gravity act perpendicular to the direction of motion, so neither force does work.
e) The net work done is zero.
6.6: From Eq. (6.2),
J.1022.50.15cos)m300)(N180(cos
4
Fs
6.7:
,J1062.214cos)m1075.0)(N1080.1(2cos2
936


Fs
or
J102.6
9

to
two places.
6.8: The work you do is:
)
ˆ
)m0.3(

mg
The displacement is upward and the gravity force is downward, so it does negative
work.
6.10: a) From Eq. (6.6),
J.1054.1
h/km
s/m
6.3
1
)km/h0.50()kg1600(
2
1
5
2

















2
2
1
 mvKE
Running:
J140)s/m2)(kg70(
2
2
1
KE
(b) Estimate:
s/m30s/ft88mph60 v

kg2000m

J109)s/m30)(kg2000(
52
2
1
KE
(c)
mghWKE 
gravity
Estimate
m2h

J20)m2)(s/m8.9)(kg1(
2
KE
6.13: Let point 1 be at the bottom of the incline and let point 2 be at the skier.

v
gives

)tan/1(2
k0

 ghv
6.14: (a)
)m0.15)(s/m80.9(2)s/m0.25(
2
2
1
2
1
22
2
f0
2
0
2
f




ghvv
mvmvmgh
KEW
s/m3.30
(b)



, down incline; displacement
sinh

 
, down incline

mghhmgW  )sin/)(sin(
||

perpendicular to incline: no displacement in this
direction, so
0

W
.

mghWWW
mg

||
, same as falling height h.
b)
tot 2 1
W K K 
gives
2
1
2

BAAB
vvmm 
. a) Solving for the ratio of the
speeds,
2/ 
BA
vv
. b) The boats are said to start from rest, so the elapsed time is the
distance divided by the average speed. The ratio of the average speeds is the same as the
ratio of the final speeds, so the ratio of the elapsed times is
2// 
BAAB
vvtt
.
6.19: a) From Eq. (6.5),
16/
12
KK 
, and from Eq. (6.6),
1
)16/15( KW 
. b) No;
kinetic energies depend on the magnitudes of velocities only.
6.20: From Equations (6.1), (6.5) and (6.6), and solving for F,
N.0.32
)m50.2(
))s/m00.4()s/m00.6)((kg00.8()(
22
2
1

F
K
s
6.22: a) If there is no work done by friction, the final kinetic energy is the work done by
the applied force, and solving for the speed,
.s/m48.4
)kg30.4(
)m20.1)(N0.36(222

m
Fs
m
W
v
b) The net work is
smgFsfFs )(
kk


, so
)kg30.4(
)m20.1))(s/m80.9)(kg30.4)(30.0(N0.36(2
)(2
2
k




m

J.1176)m0.25)(s/m80.9)(kg80.4(
2
 mgsFsW

b) Since the melon is released from rest,
0
1
K
, and Eq. (6.6) gives
J.1176
2
 WKK
6.25: a) Combining Equations (6.5) and (6.6) and solving for
2
v
algebraically,
.s/m96.4
)kg00.7(
)m0.3)(N0.10(2
)s/m00.4(2
2
tot
2
12

m
W
vv
Keeping extra figures in the intermediate calculations, the acceleration is
.s/m429.1)kg00.7/()s/mkg0.10(

plane makes with the horizontal. Using the given numbers,
s./m97.29.36sin)m75.0)(s/m80.9(2
2
v
6.27: a) The friction force is
mg
k

, which is directed against the car’s motion, so the net
work done is
mgs
k


. The change in kinetic energy is
2
01
)2/1( mvKK 
, and so
gvs
k
2
0
2/


. b) From the result of part (a), the stopping distance is proportional to the
square of the initial speed, and so for an initial speed of 60 km/h,
m3.51)0.80/0.60)(m2.91(
2






,
so that stretching 0.015 m requires
J360.0)050.0/015.0)(J4(
2

and compressing 0.020
m requires
J64.0)050.0/020.0)(J4(
2

. Another is to find the spring constant
m/N1020.3)m050.0()N160(
3
k
, from which
23
)m015.0)(m/N1020.3)(2/1( 
J360.0
and
J64.0)m020.0)(m/N1020.3)(2/1(
23

.
6.30: The work can be found by finding the area under the graph, being careful of the
sign of the force. The area under each triangle is 1/2

9.6
0
9.6
0
9.6
0



9.6
0
29.6
0
|)2/)(
m
N
0.3(|)N0.20(
xx 

J209orJ4.209mN4.71mN138 
The work is negative because the cow continues to advance as you vainly attempt to push
her backward.
6.33:
12tot
KKW 

0,
2
1
2

 ..k
, giving the same results.
6.35: a) The static friction force would need to be equal in magnitude to the spring force,
kdmg 
s
or
76.1
)s/m80.9)(kg100.0(
)m086.0)(m/N0.20(
s
2
μ
, which is quite large. (Keeping extra figures in
the intermediate calculation for d gives a different answer.) b) In Example 6.6, the
relation
2
1
2
k
2
1
2
1
mvkdmgd 

was obtained, and d was found in terms of the known initial speed
1
v
. In this case, the
condition on d is that the static friction force at maximum extension just balances the
































b) The work-energy theorem gives
s./m18.0
)kg0.4(
)J06.0(2
2

m
W
v
6.37: The work done in any interval is the area under the curve, easily calculated when
the areas are unions of triangles and rectangles. a) The area under the trapezoid is
J0.4mN0.4 
. b) No force is applied in this interval, so the work done is zero. c)
The area of the triangle is
J0.1mN0.1 
, and since the curve is below the axis
)0( 
x
F
, the work is negative, or
J0.1
. d) The net work is the sum of the results of
parts (a), (b) and (c), 3.0 J. (e)
J0.1J02J0.1  .
.
6.38: a)
J0.4K
, so
s/m00.2)kg0.2()J0.4(22  mKv
. b) No work is

1
 xx
m
k
v
6.40: a) From Eq. (6.14), with

Rddl 
,
.sin2cos2cos
0
0
02
1


wRdwRdlFW
P
P


In an equivalent geometric treatment, when
F

is horizontal,
Fdxd  lF


, and the total
work is

0
θθ
wR
θwR




6.41: a) The initial and final (at the maximum distance) kinetic energy is zero, so the
positive work done by the spring,
2
)2/1( kx
, must be the opposite of the negative work
done by gravity,
θmgLsin
, or
cm.7.5
)m/N640(
0.40sin)m80.1)(s/m80.9)(kg0900.0(2sin2
2



k
θmgL
x
At this point the glider is no longer in contact with the spring. b) The intermediate
calculation of the initial compression can be avoided by considering that between the
point 0.80 m from the launch to the maximum distance, gravity does a negative amount
of work given by

:


22
k
s/m96.1)s/m80.9)(200.0(  gμa

atvv 
0

t)s/m96.1(s/m00.80
2


s08.4t

t
mv
t
KE
P
2
2
1


W157
s08.4
)s/m00.8)(kg0.20(
2

.km800m100.8
m/W100.1)40.0(
W102.3
228
23
11



6.47: The power is
vFP 
. F is the weight, mg, so
kW.15.17s)m(2.5)sm(9.8kg)700(
2
P
So,
0.23,kW.75kW15.17 
or about
23% of the engine power is used in climbing.
6.48: a) The number per minute would be the average power divided by the work (mgh)
required to lift one box,
s,41.1
m)(0.90)sm(9.80kg)30(
hp)W(746hp)50.0(
2

or
min.6.84
b) Similarly,
s,378.0

6
ave

v
P
F
6.52: Here, Eq. (6.19) is the most direct. Gravity is doing negative work, so the rope
must do positive work to lift the skiers. The force
F

is gravity, and
,NmgF 
where N
is the number of skiers on the rope. The power is then
W.1096.2
)0.15(90.0cos
hkm6.3
sm1
h)km(12.0)sm(9.80kg)(70)50(
cos)()(
4
2







