Tài liệu Physics exercises_solution: Chapter 06 docx

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

. The work is positive, since the worker
is pushing in the same direction as the crate’s motion.

c) Since f and s are oppositely directed, Eq. (6.2) gives
J.331)m5.4)(N5.73(  fs

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.
6.5: a) See Exercise 5.37. The needed force is
N,2.99
30sin)25.0(30cos
)s/m80.9)(kg30)(25.0(
sincos
2
k
k







mg
F
keeping extra figures. b)
J5.38630cos)m50.4)(N2.99(cos 

Fs
, again keeping an

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(

ˆ
)m0.9(()
ˆ
)N40(
ˆ
)N30(( jijisF 



)m0.3)(N40()m0.9)(N30( 

J150mN120mN270 
6.9: a) (i) Tension force is always perpendicular to the displacement and does no work.
(ii) Work done by gravity is
).(
12
yymg 
When
21
yy 
,
0
mg
W
.
b) (i) Tension does no work.
(ii) Let l be the length of the string.
J1.25)2()(
12
 lmgyymgW

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

















K
b) Equation (6.5) gives the explicit dependence of kinetic energy on speed; doubling the
speed of any object increases the kinetic energy by a factor of four.
6.11: For the T-Rex,
J1032.4))hr/km4)((kg7000(
32
km/hr6.3
s/m1
2
1
K
. The person’s
velocity would be
m/s111kgJ)/701032.4(2
3
.v 
, or about 40 km/h.
6.12: (a) Estimate:
1m sv  
(walking)

2m sv  
(running)

70kgm 
Walking:
J35)s/m1)(kg70(
2
2
1

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.

12tot
KKW 

0,
2
1
2
2
01
 KmvK
Work is done by gravity and friction, so
fmg
WWW 
tot
.

mghyymgW
mg
 )(
12


tan/)sin/)(cos(
kk
mghhmgfsW
f

Substituting these expressions into the work-energy theorem and solving for
0

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)

)s/ms80.9(2
0)s/m3.30(
g2
2
1
2
1
2
222
f
2
0
2
0
2
f






vv
h
mvmvmgh
KEW

m8.46
6.15: a) parallel to incline: force component
sinmg



, 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

mgh mv
and
ghv 2
, same as if had been dropped
from height h. The work done by gravity depends only on the vertical displacement of the
object.
When the slope angle is small, there is a small force component in the direction of
the displacement but a large displacement in this direction. When the slope angle is large,
the force component in the direction of the displacement along the incline is larger but
the displacement in this direction is smaller.
c)
m0.15h
, so
s1.172  ghv
6.16: Doubling the speed increases the kinetic energy, and hence the magnitude of the
work done by friction, by a factor of four. With the stopping force given as being
independent of speed, the distance must also increase by a factor of four.
6.17: Barring a balk, the initial kinetic energy of the ball is zero, and so
J.74.2m/s)kg)(32.0145.0)(2/1()2/1(
22
 mvW
6.18: As the example explains, the boats have the same kinetic energy K at the finish line,
so
22
)2/1()2/1(
BBAA
vmvm 
, or, with
22
2,2

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

2
1
2
2
2
1







s
vvm
s
K
F
6.21:
cm8.16
)N0.40(
))s/m00.2()s/m00.6)((kg420.0(
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

smgF
v


s./m61.3
(Note that even though the coefficient of friction is known to only two places, the
difference of the forces is still known to three places.)
6.23: a) On the way up, gravity is opposed to the direction of motion, and so
J4.28)m0.20)(s/m80.9)(kg145.0(
2
 mgsW
.
b)
s/m26.15
)kg145.0(
)J4.28(2
)s/m0.25(2
22
12



m
W
vv
.
c) No; in the absence of air resistance, the ball will have the same speed on the way
down as on the way up. On the way down, gravity will have done both negative and
positive work on the ball, but the net work will be the same.
6.24: a) Gravity acts in the same direction as the watermelon’s motion, so Eq. (6.1) gives

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(

22
a
From Eq. (2.13), with appropriate change in
notation,
),m0.3)(s/m429.1(2)s/m00.4(2
222
1
2
2
 asvv
giving the same result.
6.26: The normal force does no work. The work-energy theorem, along with Eq. (6.5),
gives
,sin22
22

gLgh
m
W
m
K
v

where
sinh L


is the vertical distance the block has dropped, and

is the angle the

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

s
. (This method avoids the intermediate calculation of
k

, which in this case is about 0.279.)
6.28: The intermediate calculation of the spring constant may be avoided by using Eq.
(6.9) to see that the work is proportional to the square of the extension; the work needed
to compress the spring 4.00 cm is
 
J3.21)J0.12(
2
cm00.3
cm00.4

.
6.29: a) The magnitude of the force is proportional to the magnitude of the extension or
compression;
N.64)m050.0/m020.0)(N160(,N48)m050.0/m015.0)(N160( 
b) There are many equivalent ways to do the necessary algebra. One way is to note
that to stretch the spring the original 0.050 m requires
J4)m050.0(
m0.050
N169
2
2
1










,
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

base height
.
a)
J40)N10)(m8(2/1 
.
b)
J20)N10)(m4(2/1 
.
c)
J60)N10)(m12(2/1 
.
6.31: Use the Work-Energy Theorem and the results of Problem 6.30.
a)
s/m83.2
kg10
)J40)(2(
v
b) At
m12x
, the 40 Joules of kinetic energy will have been increased by 20 J, so
s/m46.3
kg10
)J60)(2(
v
.
6.32: The work you do with your changing force is
xdxdxdxxF
m
N
0.3)N0.20()(

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

2
01
 KmvK
Work is done by the spring force.
2
2
1
tot
kxW 
, where x is the amount the spring is
compressed.
cm5.8/and
2
1
2
1
0
2
0
2
 kmvxmvkx
6.34: a) The average force is
N400)m200.0/()J0.80( 
, and the force needed to hold
the platform in place is twice this, or 800 N. b) From Eq. (6.9), doubling the distance
quadruples the work so an extra 240 J of work must be done. The maximum force is
quadrupled, 1600 N.
Both parts may of course be done by solving for the spring constant
N/m10004m)2000(J)0.80(2
32

 ..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

spring force, or
mgkd
s


. Solving for
2
1
v
and substituting,
)),47.0)(60.0(2)60.0((
)m/N0.20(
)s/m80.9)(kg10.0(
)2(
2
2
2
22
ks
2
s
2
s
k
2
s
k
22
1


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

k
mg
k
mg
g
k
mg
m
k
dgdd
m
k
v
from which
s/m67.0
1
v
.
6.36: a) The spring is pushing on the block in its direction of motion, so the work is
positive, and equal to the work done in compressing the spring. From either Eq. (6.9) or
Eq. (6.10),
J06.0)m025.0)(m/N200(
2
2
1
2
2
1
 kxW
.

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

done between
m0.3x
and
m0.4x
, so the speed is the same, 2.00 m/s. c)
J0.3K
, so
s/m73.1)kg0.2/()J0.3(2/2  mKv
.
6.39: a) The spring does positive work on the sled and rider;
22
)2/1()2/1( mvkx 
, or
s/m83.2)kg70/()m/N4000()m375.0(/  mkxv
. b) The net work done by
the spring is
)()2/1(
2
2
2
1
xxk 
, so the final speed is
s./m40.2))m200.0()m375.0((
)kg70(
m/N4000(
)(
222
2
2

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

wF 2
times the horizontal distance, in this case (see Fig. 6.20(a))
0
sinθR
,
giving the same result. b) The ratio of the forces is
0
tan
2
cot2
0
θ
w
w


.
c)
.
2
cot2
)cos1(
sin
2
)cos1(
sin2
0
0
0
0

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

J567.00.40sin)m80.0m80.1)(s/m80.9)(kg0900.0(
2

, and so
the kinetic energy of the glider at this point is 0.567 J. At this point the glider is no longer
in contact with the spring.
6.42: The initial and final kinetic energies of the brick are both zero, so the net work done
on the brick by the spring and gravity is zero, so
0)21(
2
 mghkd
, or
m.53.0)m/N450/()m6.3)(s/m80.9)(kg80.1(2/2
2
 kmghd
The spring will
provide an upward force while the spring and the brick are in contact. When this force
goes to zero, the spring is at its uncompressed length.
6.43:
J106.3)s3600)(W100()time)(power(Energy
5

kg.70fors1002so
2
1
2
 mK/mvmvK
6.44: Set time to stop:
mamgmaF 
k

:


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

2
1

6.45: The total power is
W10485.1)s/m00.9)(N165(
3

, so the power per rider is
742.5 W, or about 1.0 hp (which is a very large output, and cannot be sustained for long
periods).
6.46: a)
W.102.3
yr)/s1016.3(
)yr/J100.1(
11
7
19



b)
.kW/person2.1
folks106.2
W102.3
8
11



c)

.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

m)(0.90)sm(9.80kg)(30
W)100(
2

or
min.7.22
6.49: The total mass that can be raised is
kg,2436
m)(20.0)sm(9.80
s)hp)(16.0W(746hp)0.40(
2

so the maximum number of passengers is
.28
kg65.0
kg1836

6.50: From any of Equations (6.15), (6.16), (6.18) or (6.19),
hp.3.57 W1066.2
s)(4.00
m)(2.80N)3800(
3

t
Wh
P
6.51:
N.101.8
s))m6.3(h)km1((h)km65(
hp)Whp)(746(280,000)70.0()70.0(

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














vNmgP
Note that Eq. (1.18) uses

as the angle between the force and velocity vectors; in this
case, the force is vertical, but the angle
0.15
is measured from the horizontal, so
 0.150.90

is used.