Download Unit 6 Work and Energy Solutions to HW 1 and 2

1.
The minimum force required to lift the firefighter is equal to his weight. The force and the
displacement are both upward, so the angle between them is 0. Use Eq. 6–1.
Wclimb  Fclimb d cos   mgd cos   (75.0 kg)(9.80 m/s 2 )(28.0 m) cos 0  2.06 10 4 J
2.
The maximum amount of work would be the work done by gravity. Both the force and the
displacement are downward, so the angle between them is 0. Use Eq. 6–1.
WG  mgd cos  (1.2 kg)(9.80 m/s2 )(0.50 m)cos 0  5.9 J
This is a small amount of energy. If the person adds a larger force to the hammer during the fall,
then the hammer will have a larger amount of energy to give to the nail.
3.
Draw a free-body diagram for the crate as it is being pushed across the
floor. Since it is not accelerating vertically, FN  mg. Since it is not
accelerating horizontally, FP  Ffr  k FN  k mg. The work done to
move it across the floor is the work done by the pushing force. The angle
between the pushing force and the direction of motion is 0.
x
FP
Ffr
mg
FN
Wpush  Fpush d cos0  k mgd (1)  (0.50)(46.0 kg)(9.80 m/s2 )(10.3 m)  2300 J
4.
(a)
x
See the free-body diagram for the crate as it is being pulled. Since the
crate is not accelerating horizontally, FP  Ffr  230 N. The work done
to move it across the floor is the work done by the pulling force. The
angle between the pulling force and the direction of motion is 0. Use
Eq. 6–1.
Ffr
FP
mg
FN
WP  FP d cos 0  (230 N)(50 m)(1)  1150 J  1200 J
(b)
See the free-body diagram for the crate as it is being lifted. Since the crate
is not accelerating vertically, the pulling force is the same magnitude as the
weight. The angle between the pulling force and the direction of motion is 0°.
y
FP
WP  FP d cos 0  mgd  (1200 N)(5.0 m)  6.0  103 J
5.
Draw a free-body diagram of the car on the incline. The minimum work
will occur when the car is moved at a constant velocity. Write Newton’s
second law in the x direction, noting that the car is not accelerated. Only
the forces parallel to the plane do work.
 Fx  FP  mg sin  0  FP  mg sin 
The work done by FP in moving the car a distance d along the plane
mg
y
FN
FP



mg
(parallel to FP ) is given by Eq. 6–1.
WP  FP d cos 0  mgd sin   (950 kg)(9.80 m/s 2 )(710 m) sin 9.0  1.0  106 J
6.
The distance over which the force acts is the area to be mowed divided by the width of the mower.
The force is parallel to the displacement, so the angle between them is 0 Use Eq. 6–1.
W  Fd cos   F
A
200 m2
cos   (15 N)
 6000 J
w
0.50 m
x
7.
The minimum work required to shelve a book is equal to the
weight of the book times the vertical distance the book is
moved. See the diagram. Each book that is placed on the
lowest shelf has its center of mass moved upward by
15.0 cm 11.0 cm  26.0 cm. So the work done to move
28 books to the lowest shelf is W1  28mg (0.260 m). Each
book that is placed on the second shelf has its center of mass
26.0 cm
moved upward by 15.0 cm  38.0 cm  11.0 cm  64.0 cm,
so the work done to move 28 books to the second shelf is
W2  28mg (0640 m). Similarly, W3  28mg (1020 m), W4  28mg (1400 m), and
W5  28mg (1.780 m). The total work done is the sum of the five work expressions.
W  28mg (0.260 m  0.640 m  1.020 m  1.400 m  1.780 m)
 28(1.40 kg)(9.80 m/s2 )(5.100 m)  1959 J  1960 J
3rd shelf
2nd shelf
64.0 cm
1st shelf
floor
8.
Consider the diagram shown. If we assume that the man pushes
straight down on the end of the lever, then the work done by
the man (the “input” work) is given by WI  FI hI . The object
moves a shorter distance, as seen from the diagram, so
WO  FO hO . Equate the two amounts of work.
WO  WI
 FO hO  FI hI
But by similar triangles, we see that
9.

hI

hO
FI
hI
I
FO hI

FI hO
I
, so
O
hO
O
FO

FI
I
FO
.
O
Since the acceleration of the box is constant, use Eq. 2–11b to find the distance moved. Assume that
the box starts from rest.
d  x  x0  0t  12 at 2  0  12 (2.0 m/s2 )(7.0 s)2  49 m
Then the work done in moving the crate is found using Eq. 6–1.
W  Fd cos 0  mad  (4.0 kg)(2.0 m/s2 )(49 m)  390 J
10.
The piano is moving with a constant velocity down the plane. FP is the force
of the man pushing on the piano.
(a)
Write Newton’s second law on each direction for the piano, with an
acceleration of 0.
 Fy  FN  mg cos  0
 Fx  mg sin   FP  0
 FN  mg cos 
FN
y
FP

mg


FP  mg sin   mg sin 
 (380 kg)(9.80 m/s 2 )(sin 25)  1574 N  1600 N
(b)
The work done by the man is the work done by FP  The angle between FP and the direction
of motion is 180 Use Eq. 6–1.
WP  FP d cos180  (1574 N)(29 m)  4565 J  4600 J
(c)
The angle between the force of gravity and the direction of motion is 65. Calculate the work
done by gravity.
WG  FG d cos 63  mgd cos 63  (380 kg)(9.80 m/s 2 )(2.9 m) cos 65
 4564 N  4600 J
(d)
Since the piano is not accelerating, the net force on the piano is 0, so the net work done on the
piano is also 0. This can also be seen by adding the two work amounts calculated.
Wnet  WP  WG  4600 J  4600 J  0
11.
If the person pulls 2 m of rope through his hands, the rope holding the piano will get shorter by 2 m.
But that means the rope on the right side of the pulley will get shorter by 1 m, and the rope on the
left side will also get shorter by 1 m. Thus for each meter the load is raised, 2 m of rope must be
pulled up.
In terms of energy (assuming that no work is lost to friction), the work done by the man pulling on
the rope must be equal to the work done on the piano. If the piano has weight mg, and it moves
x
upward a distance d, then the work done on the piano is mgd. The person pulls the rope a distance
2d, and therefore must exert a force of ½mg to do the same amount of work.
Wdone by  Wdone on
man
12.
 Fpull (2d )  mgd
piano
 Fpull  12 mg
Consider a free-body diagram for the grocery cart being pushed up
the ramp. If the cart is not accelerating, then the net force is 0 in all
directions. This can be used to find the size of the pushing force. The
angles are   17 and   12. The displacement is in the x
direction. The work done by the normal force is 0 since the normal
force is perpendicular to the displacement. The angle between the
force of gravity and the displacement is 90    102. The angle
between the normal force and the displacement is 90. The angle
between the pushing force and the displacement is     29.
 Fx  FP cos (   )  mg sin   0
 FP 
y
FN

FP


mg
mg sin 
cos (   )
Wmg  mgd cos 112  (16 kg)(9.80 m/s 2 )(7.5 m) cos 102  244.5 J  240 J
Wnormal  FN d cos 90  0
 mg sin 12 
WP  FP d cos 29  
 d cos 29  mgd sin 12
 cos 29 
 (16 kg)(9.80 m/s 2 )(7.5 m) sin 12  244.5 J  240 J
13.
The work done will be the area under the Fx vs. x graph.
(a)
From x  0.0 to x  10.0 m, the shape under the graph is trapezoidal. The area is
W  (400 N) 12 (10 m  4 m)  2800 J
(b)
From x  10.0 m to x  15.0 m, the force is in the opposite direction from the direction of
motion, so the work will be negative. Again, since the shape is trapezoidal, we find
W  ( 200 N) 12 (5 m  2 m)  700 J
Thus the total work from x  00 to x  150m is 2800 J  700 J  2100 J 
14.
(a)
The gases exert a force on the jet in the same direction as the displacement of the jet. From
the graph we see the displacement of the jet during launch is 85 m. Use Eq. 6–1 to find the
work.
Wgas  Fgas d cos 0  (130  103 N)(85 m)  1.1107 J
(b)
The work done by catapult is the area underneath the graph in Fig. 6–39b. That area is a
trapezoid.
Wcatapult  12 (1100  103 N  65  103 N)(85 m)  5.0  107 J
15.
Find the velocity from the kinetic energy, using Eq. 6–3.
KE
 12 m 2
 
2KE

m
2(6.21 1021 J)
5.31 1026
 484 m/s
x
16.
(a)
Since
KE
 12 m 2 ,   2KE/m and  
will be multiplied by a factor of
(b)
Since
KE
KE .
Thus if the kinetic energy is tripled, the speed
3.
 12 m 2 , KE   2 . Thus if the speed is halved, the kinetic energy will be multiplied
by a factor of 1/4 .
17.
The work done on the electron is equal to the change in its kinetic energy, Eq. 6–4.
W  KE  12 m22  12 m12  0  12 (9.11 1031 kg)(1.10  106 m/s) 2  5.5110 19 J
Note that the work is negative since the electron is slowing down.
18.
The work done on the car is equal to the change in its kinetic energy, Eq. 6–3.
2

 1 m/s  
5
W  KE  12 m22  12 m12  0  12 (925 kg) (95 km/h) 
   3.2 10 J
3.6
km/h



Note that the work is negative since the car is slowing down.
19.
The kinetic energies of both bullets are the same. Bullet 1 is the heavier bullet.
m1  2m2
1 m 2
2 2 2
 12 m112
 m222  2m212
 22  212
 2  1 2
The lighter bullet has the higher speed, by a factor of the square root of 2. Both bullets can do the
same amount of work.
20.
The force of the ball on the glove will be the opposite of the force of the glove on the ball, by
Newton’s third law. The objects have the same displacement, so the work done on the glove is
opposite the work done on the ball. The work done on the ball is equal to the change in the kinetic
energy of the ball, Eq. 6–4.
Won ball  (KE2  KE1 )ball  12 m22  12 m12  0  12 (0.145 kg)(32 m/s)2  74.24 J
So Won glove  74.24 J. But Won glove  Fon glove d cos0, because the force on the glove is in the
same direction as the motion of the glove.
74.24 J  Fon glove (0.25 m)  Fon glove 
74.24 J
 3.0 102 N , in the direction of the
0.25 m
original velocity of the ball.
21.
The force exerted by the bow on the arrow is in the same direction as the displacement of the arrow.
Thus W  Fd cos 0  Fd  (105 N)(0.75 m)  78.75 J. But that work changes the kinetic energy of
the arrow, by the work-energy theorem. Thus
Fd  W  KE 2  KE1  12 m22  12 m12
22.
 2 
The work needed to stop the car is equal to the change in
the car’s kinetic energy. That work comes from the force
of friction on the car. Assume the maximum possible
frictional force, which results in the minimum braking
distance. Thus Ffr  s FN . The normal force is equal to
the car’s weight if it is on a level surface, so Ffr  s mg.
In the diagram, the car is traveling to the right.
2Fd
2(78.75 J)
 12 
 0  43 m/s
m
0.085 kg
d = stopping distance
Ffr
FN
mg
W  KE  Ffr d cos180  12 m22  12 m12
  s mgd   12 m12
 d
12
2g s
Since d  12 , if 1 increases by 50%, or is multiplied by 1.5, then d will be multiplied by a factor
of (1.5) 2 , or 2.25.