Chapter 7
KINETIC ENERGY AND WORK
Problems for chapter 7
W
energy of the trunk increase or decrease?
Problem 14.
Fnet d
ϕ2 = 180o − 60o
=120o
(3.82 N) (4.00 m) 15.3 J
F2
d
Fig. 7-26
Motion
••15
Figure 7-27 shows three forces applied to a trunk that
θ direction
G G floor. The force magnimoves leftward by 3.00 m over a frictionless
F1
where
used
fact that d || Fnet (which follows from the fact that the canister
θ
tudes
areweF1have
! 5.00
N, Fthe
2 ! 9.00 N, and F3 ! 3.00 N, and the indistarted
from
and During
movedthe
horizontally
the action
of horizontal forces — the
cated
angle
is u rest
! 60.0°.
displacement,
(a) what
is the net
G under
Fig. 7-29 P
work
done effect
on theof
trunk
by is
the
three forces
resultant
which
expressed
by and
Fnet ).(b) does the kinetic
F3
energy
276 of the trunk increase or decrease?
CHAPTER
7
••20 A block is sent up a friction
Fig.is
7-27
Problem
15. (a) The forces are constant, so the work done by any one of
them
given
by 15.
extends
upward.
Figure
7-30 gives
G
F2
G G
G
of athe
displacement,
W276F ˜ d , where d is the displacement. Force F1 is in the direction as
CHAPTER
7 x; the sca
function
of position
••16
An 8.0 kg object is moving in the positive dire
by Kthrough
block’s forc
ini
G
so
θ
s ! 40.0x J.
x axis. When itset
passes
!If0,the
a constant
F1angle
Force F2 makes anW
withN)
the(3.00
displacement,
so
normal
block?
along
axisJ.begins
toforce
act onon
it. the
Figure
7-28 gives its
F1d of
cos120°
I1 (5.00
m) cos
0q the
15.0
1
G
Force F2 makes an angle of 120° with the displacement, so
F2 d cos I2
W2
K (J)
K0
d cos IF3
G
Force F3 is perpendicular to the displacement, so
d cos
I3 = 0direction
since cos
90°
W3 = F
••16
An 8.0 kg object is moving
in3the
positive
of an
F3d0,cos
I3 = 0 since
cosdirected
90° = 0.
W3 =
x axis. When it passes through
x!
a constant
force
along
begins
onthree
it. Figure
7-28isgives its kinetic enThe the
net axis
work
donetobyactthe
forces
The net work done by the three forces is
K0
W
W
W W W
= 0.
K ( J)
W2 F2
(9.00 N) (3.00 m) cos120q 13.5 J.
G
2
Force F3 is perpendicular
to the displacement, so
Fig. 7-27 Problem 15.
K (J)
Ks
(9.00 N) (3.00 m) cos120q 13.5 J.
0
Fig. 7-28
5
x (m)
0
Problem 16.
Fig. 7-30
x(
P
** View A
15.0 J 13.5 J 0 ••21
1.50 SSM
J. A cord is used to vertic
1
W1 W2 2 W3 315.0 J 13.5 J 0 1.50 block
J.
of mass M at a constant down
the block has fallen a distance d,
(b)
dowork
workononthethe
box,
its kinetic
energy
increases
byJ on
1.50
during(b)
thethe
(b)IfIfno
noother
other forces
forces do
box,
its kinetic
energy
increases
by 1.50
during
cord’s
force
theJ the
block,
displacement.
force on the block, (c) the kinetic
displacement.
't 8.0 s
with a speed of (b) 1.0 m/s and (c) 2.0 m/s?
has a mass of 2
speed
at the
(c)
Since
the speed
of100
2.0kg
m/sblock
is constant,
8.0atmeters
is traveled
Using
Eq. end
SSM
ILW A
•45
is pulled
a constant
speedinof4.0
5.0seconds.
7-42,
with average
power floor
replaced
by power,
haveof 122 N directed
m/s across
a horizontal
by an
appliedwe
force
37° above the horizontal. What is the rate at which the force does
W 900 J
F2
work on the block?
P
= 225 W | 2.3 u 102 W .
t 4.0 s has a mass of 3.0 ! 103 kg and
•46 The loaded cab of an'elevator
moves 210 m up the shaft in 23 s atGconstant speed. AtGwhat
G aver- G
F
is
given
by
P
F
˜
v
45.
The
power
associated
with
force
age rate does the force from the cable do work on the cab? , where v is the velocity
of the object on which the force acts. Thus,
••47 A machine carries a 4.0 kg package from an initial position
:
G G$ (0.75 m)ĵ $ (0.20 m)k̂ at t " 0 to a final posi-2
of di " (0.50 m)î
F ˜ v Fv cos I (122 N)(5.0 m/s)cos37q 4.9 u 10 W.
:P
tion of df " (7.50 m)î $ (12.0 m)ĵ $ (7.20 m)k̂ at t " 12 s. The
54 The only f
constant force applied by the machine on the package is
kgthere
bodyisasnothe b
: Recognizing that the force in the cable must equal the total weight (since
46.
F " (2.00 N)î $ (4.00 N)ĵ $ (6.00 N)k̂ . For that displacement,
x axis varies as
acceleration), we employ Eq. 7-47:
find (a) the work done on the package by the machine’s force and
The scale of the
'x
(b) the average power of the P
machine’s
force
on
the
package.
Fv cos T
mg
is set by Fs " 4.0
't
••48 A 0.30 kg ladle sliding on a horizontal frictionless surface is
body at x " 0 is
attached
to one
endtheoffact
a horizontal
500 of
N/m)
theelevator’s
kinetic energ
where
we have
used
that T 0qspring
(both (k
the"force
the whose
cable and the
other end
is fixed. The
ladle has a kinetic energy of 10 J as it passes
3.0 m? (b) At wh
motion
are upward).
Thus,
through its equilibrium position (the point210
at m
which the spring
body have a kin
·
3
2 §
5
u10
) ¨ doing ¸work
2.7on
u10theW.
force is zero). (a)PAt (3.0
what
ratekg)(9.8
is the m/s
spring
la(c) What is the m
23 s ¹
©
dle as the ladle passes through its equilibrium position? (b) At
and x " 5.0 m?
what rate is the spring doing work on the ladle when the spring is
55 SSM A hor
47.
(a) Equation
7-8myields
compressed
0.10
and the ladle is moving away from the equilibabove the horizo
!# $&
" %
Chapter 9
CENTER OF MASS AND LINEAR
MOMENTUM
sec. 9-5 The Linear Momentum
Boat's displacement db
of a System of Particles
(b)
•18 A 0.70 kg ball moving horizontally at 5.0 m/s strikes a vertical
Fig. 9-45 Problem 17.
wall and rebounds with speed 2.0
m/s. What is the magnitude of the change in its linear momentum?
•19 ILW A 2100 kg truck traveling north at 41 km/h turns east and
accelerates to 51 km/h. (a) What is the change in the truck’s kinetic
!p = m v f " vi = 0.7 "2 " 5 = 0.7 "7 = 4.9kg.m / s
energy? What are
the (b) magnitude and (c) direction of the
change in its momentum?
••20
At time t ! 0, a ball is struck at ground level and sent
over level ground. The momentum p versus t during the flight is
tions Here **