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Unformatted text preview: CHAPTER 5
Quick Quizzes
1. (c), (a), (d), (b). The work in (c) is positive and of the largest possible value because the
angle between the force and the displacement is zero. The work done in (a) is zero because
the force is perpendicular to the displacement. In (d) and (b), negative work is done by the
applied force because in neither case is there a component of the force in the direction of
the displacement. Situation (b) is the most negative value because the angle between the
force and the displacement is 180°. 2. All three balls have the same speed the moment they hit the ground because all start with
the same kinetic energy and undergo the same change in gravitational potential energy. 3. (c). 4. (c). The decrease in mechanical energy of the system is fk∆x. This is smaller than the value
on the horizontal surface for two reasons: (1) the force of kinetic friction fk is smaller
because the normal force is smaller, and (2) the displacement ∆x is smaller because a
component of the gravitational force is pulling on the book in the direction opposite to its
velocity. 135 CHAPTER 5 Problem Solutions
5.1 If the weights are to move at constant velocity, the net force on them must be zero. Thus,
the force exerted on the weights is upward, parallel to the displacement, with
magnitude 350 N. The work done by this force is
W = ( F co s θ ) s = ⎡( 350 N ) cos 0°⎤( 2.00 m ) = 700 J .
⎣
⎦ 5.2 To lift the bucket at constant speed, the woman exerts an upward force whose
magnitude is F = mg = ( 20.0 kg ) 9.80 m s 2 = 196 N . The work done is W = ( F co s θ ) s , so ( ) the displacement is s= W
6.00 × 103 J
=
= 30.6 m .
F cos θ (196 N ) cos 0° ( ) ( ) 5.3 W = ( F co s θ ) s = ⎡ 5.00 × 10 3 N co s 0°⎤ 3.00 × 10 3 m = 1.50 × 107 J = 15.0 M J .
⎣
⎦ 5.4 The applied force makes an angle of 25° with the displacement of the cart. Thus, the
work done on the cart is
W = ( F co s θ ) s = ⎡( 35 N ) co s 25°⎤( 50 m ) = 1.6 × 10 3 J = 1.6 k J .
⎣
⎦ 5.5 ( ) (a) The force of gravity is given by mg = ( 5.00 kg ) 9.80 m s 2 = 49.0 N and is directed
downwards. The angle between the force of gravity and the direction of motion is θ = 90.0°  30.0° = 60.0°, and so the work done by gravity is given as
W g = ( F co s θ ) s = ⎡( 49.0 N ) co s 60.0°⎤( 2.50 m ) = 61.3 J .
⎣
⎦ (b) The normal force exerted on the block by the incline is n = mg co s 30.0° , so the
friction force is fk = µk n = ( 0.436) ( 49.0 N ) cos 30.0° = 18.5 N .
This force is directed opposite to the displacement (i.e. θ = 180°), and the work it
does is
W f = ( fk co s θ ) s = [ (18.5 N ) co s 180°] ( 2 .50 m ) = − 46.3 J . (c) Since the normal force is perpendicular to the displacement; θ = 90°, cos θ = 0 , and
the work done by the normal force is zero .
136 CHAPTER 5.6 5 The total distance the scraper is moved over the surface of the tooth is
s = 20 ( 0.75 cm ) = 15 cm = 0.15 m . The friction force has magnitude
fk = µk n = ( 0.90) ( 5.0 N ) = 4.5 N . Hence, the force which must be applied in the direction
of the motion to overcome friction is F = 4.5 N and the work done is
W = ( F co s θ ) s = ⎡( 4.5 N ) co s 0°⎤( 0.15 m ) = 0.68 J .
⎣
⎦ 5.7 (a) ΣFy = F sin θ + n − mg = 0 s = 20.0 m n = m g − F s in θ n ΣFx = F co s θ − µ k n = 0 n= ∴ fk = µkn F co s θ 18.0 kg F
θ = 20.0° µk
mg − F sin θ = mg F co s θ µk ( 0.500) (18.0 kg ) ( 9.80 m s )
µk mg
F=
=
= 79.4 N
µk sin θ + cos θ ( 0.500) sin 20.0° + co s 20.0°
2 (b) W F = ( F co s θ ) s = ⎡( 79.4 N ) cos 20.0°⎤( 20.0 m ) = 1.49 × 103 J = 1.49 kJ
⎣
⎦
(c) fk = F cos θ = 74.6 N
W f = ( f k co s θ ) s= ⎡( 74.6 N ) co s 180°⎤( 20.0 m ) = − 1.49 × 103 J = − 1.49 k J
⎣
⎦ 5.8 (a) W F = ( F co s θ ) s = ⎡(16.0 N ) co s 25.0°⎤( 2 .20 m )
⎣
⎦ n 16.
0N
25.0° W F = 31.9 J s = 2.20 m
2.50 kg (b) Wn = ( n co s 90°) s = 0
(c) F= W g = ( m g co s 90°) s = 0 mg (d) Wnet = WF + Wn + W g = 31.9 J + 0 + 0 = 31.9 J 137 CHAPTER 5.9 5 (a) The workenergy theorem, Wnet = KE f − KEi , gives
5000 J = ( ) 1
2 .50 × 103 k g v 2 − 0 , or
2 v = 2 .00 m s . (b) W = ( F co s θ ) s = ( F co s 0°) ( 25.0 m ) = 5000 J , so F = 200 N . 5.10 1
Requiring that KEping pong = KEbowling with KE = m v 2 , we have
2 ( ) 1
1
2
2 .45 × 10−3 k g v 2 = ( 7.00 k g ) ( 3.00 m s ) , giving v = 160 m s .
2
2 5.11 The person’s mass is m = w
700 N
=
= 71.4 k g . The net upward force acting on the
g 9.80 m s 2 body is Fnet = 2 ( 355 N ) − 700 N = 10.0 N . The final upward velocity can then be calculated
from the workenergy theorem as
1
1
Wnet = KE f − KEi = m v 2 − m v i2 ,
2
2 or ( Fnet cos θ ) s = ⎡(10.0 N ) cos 0°⎤( 0.250 m ) = 2 (71.4 kg ) v 2 − 0
⎣
⎦
1 which gives 5.12 v = 0.265 m s u p w a r d . (a) W g = ⎡mg cos θ ⎤s = ⎡mg cos ( 90.0° + φ ) ⎤ s
⎣
⎦⎣
⎦ ( ) W g = ⎡(10.0 k g ) 9.80 m s 2 co s 110°⎤ ( 5.00 m )
⎣
⎦ s= m n = mg cos φ F = 100 N 10.0 kg = − 168 J fk = µkn (b) W f = ( fk cos θ ) s = ⎡( µk n ) cos 180°⎤ s
⎣
⎦ = ⎡0.400 ( 98.0 N cos 20.0°) cos 180°⎤( 5.00 m )
⎣
⎦
= − 184 J (c) 5.00 W F = ( F co s θ ) s = ⎡(100 N ) cos 0°⎤( 5.00 m ) = 500 J
⎣
⎦ 138 φ mg φ = 20.0° CHAPTER 5 (d) ∆KE = Wnet = W g + W f + WF = 148 J
(e) 1
1
∆KE = mv 2 − mv i2
2
2 v= 5.13 2 ( ∆KE)
2 (148 J)
2
+ v i2 =
+ (1.50 m s ) = 5.64 m s
m
10.0 kg (a) We use the workenergy theorem to find the work.
1
1
1
2
2
W = ∆KE = m v 2 − m v i2 = 0 − ( 70 k g ) ( 4.0 m s ) = − 5.6 × 10 J .
2
2
2 (b) W = ( F co s θ ) s = ( fk co s 180°) s = ( − µ k n ) s = ( − µ k m g ) s ,
so 5.14 ( ) −5.6 × 10 2 J
W
=−
= 1.2 m .
s=−
µ k mg
( 0.70) ( 70 k g ) 9.80 m s 2 ( ) At the top of the arc, v y = 0 , and v x = v ix = v i co s 30.0° = 34.6 m s .
2
2
Therefore v 2 = v x + v y = ( 34.6 m s ) , and
2 1
1
2
KE = m v 2 = ( 0.150 kg ) ( 34.6 m s ) = 90.0 J .
2
2 5.15 (a) The final kinetic energy of the bullet is ( ) 1
1
2
KE f = m v 2 = 2.0 × 10−3 k g ( 300 m s ) = 90 J .
2
2 (b) We know that
Thus, F = 5.16 (a) ( ) W = ∆KE , and also W = F co s θ s . ∆KE
90 J − 0
2
=
= 1.8 × 10 N
s co s θ ( 0.50 m ) cos 0° 121
2
KEA = m v A = ( 0.60 k g ) ( 2.0 m s ) = 1.2 J
2
2 2 ( KEB )
2 ( 7.5 J)
12
=
= 5.0 m s
(b) KEB = m v B , so v B =
m
0.60 kg
2
139 CHAPTER 5 (c) Wnet = ∆KE = KEB − KEA = ( 7.5 − 1.2) J = 6.3 J
5.17 Wnet = ( Froad co s θ1 ) s + ( Fresist co s θ 2 ) s = ⎡(1000 N ) co s 0°⎤s + ⎡( 950 N ) co s 180°⎤s
⎣
⎦⎣
⎦ Wnet = (1000 N 950 N ) ( 20 m ) = 1.0 × 103 J
1
Also, Wnet = KE f − KEi = m v 2 − 0 , so
2 v= 5.18 ( ) 2 1.0 × 103 J
2Wnet
=
= 1.0 m s
m
2000 kg The initial kinetic energy of the sled is
1
1
2
KEi = m v i2 = (10 k g ) ( 2.0 m s ) = 20 J ,
2
2 and the friction force is fk = µk n = µk mg = ( 0.10) ( 98 N ) = 9.8 N .
Wnet = ( fk co s 180°) s = KE f − KEi , so s = 5.15 0 − KEi
−20 J
=
= 2.0 m
fk co s 180° −9.8 N (a) The final kinetic energy of the bullet is ( ) 1
1
2
KE f = m v 2 = 2.0 × 10−3 k g ( 300 m s ) = 90 J .
2
2 (b) We know that
Thus, F = 5.16 (a) ( ) W = ∆KE , and also W = F cos θ s . ∆KE
90 J − 0
2
=
= 1.8 × 10 N
s co s θ ( 0.50 m ) cos 0° 121
2
KEA = m v A = ( 0.60 k g ) ( 2.0 m s ) = 1.2 J
2
2 2 ( KEB )
2 ( 7.5 J)
12
=
= 5.0 m s
(b) KEB = m v B , so v B =
m
0.60 kg
2
(c) Wnet = ∆KE = KEB − KEA = ( 7.5 − 1.2) J = 6.3 J
140 CHAPTER 5.17 5 Wnet = ( Froad co s θ1 ) s + ( Fresist co s θ 2 ) s = ⎡(1000 N ) co s 0°⎤s + ⎡( 950 N ) co s 180°⎤s
⎣
⎦⎣
⎦ Wnet = (1000 N 950 N ) ( 20 m ) = 1.0 × 103 J
1
Also, Wnet = KE f − KEi = m v 2 − 0 , so
2 v= 5.18 ( ) 2 1.0 × 103 J
2Wnet
=
= 1.0 m s
m
2000 kg The initial kinetic energy of the sled is
1
1
2
KEi = m v i2 = (10 k g ) ( 2.0 m s ) = 20 J ,
2
2 and the friction force is fk = µk n = µk mg = ( 0.10) ( 98 N ) = 9.8 N .
Wnet = ( fk co s 180°) s = KE f − KEi , so s = 5.19 0 − KEi
−20 J
=
= 2.0 m
f k co s 180° −9.8 N ( ) The weight of the blood is mg = ( 0.50 kg ) 9.80 m s 2 = 4.9 N .
(a) The displacement from the feet to the heart is y = 1.3 m so the potential energy,
using the feet as the reference level, is PEg = mgy = ( 4.9 N ) (1.3 m ) = 6.4 J .
(b) The displacement from the head to the heart is y = (1.3 − 1.8) m = −0.50 m . Thus,
when the head is used as the reference level, the potential energy is PEg = mgy = ( 4.9 N ) ( −0.50 m ) = − 2.5 J .
5.20 (a) Relative to the ceiling, y = 1.00 m. ( ) Thus, PEg = mgy = ( 2.00 kg ) 9.80 m s 2 ( −1.00 m ) = − 19.6 J .
(b) Relative to the floor, y = 2.00 m, so ( ) PEg = mgy = ( 2.00 k g ) 9.80 m s 2 ( 2 .00 m ) = 39.2 J . 141 CHAPTER 5 (c) Relative to the height of the ball, y = 0, and PEg = 0 .
5.21 The work the muscle must do against the force of gravity to raise the arm is ( ) ( ) Wmuscle = −W g = − mgy i − m gy f = m g y f − y i = mgh = m gL (1 − co s 30°) , or 5.22 ( ) Wmuscle = ( 7.0 k g ) 9.80 m s 2 ( 0.21 m ) (1 − cos 30°) = 1.9 J . When the legs are straightened, the spring will be stretched by
x = ( 0.720 m ) (1 − sin 40.0°) + ( 0.740 m ) (1 − sin 25.0°) = 0.684 m . The work the legs must do to stretch the spring this amount is
1
1
2
W = PEs = kx 2 = ( 250 N m ) ( 0.684 m ) = 58.6 J .
2
2 5.23 The equivalent spring constant for the muscle is
k= F
105 N
=
= 5.25 × 103 N m .
2
x 2 .00 × 10 m The work done in stretching this by 2.00 cm is ( )( 1
1
W = PEs = kx 2 = 5.25 × 103 N m 2 .00 × 102 m
2
2 5.24 ) 2 = 1.05 J . Let m be the mass of the ball, R the radius of the circle, and F the 30.0 N force. With y = 0
at the bottom of the circle, Wnc = ( KE + PE) f − ( KE + PE)i yields ( F cos 0°) π R = ⎛ 2 mv 2f + 0⎞ − ⎛ 2 mvi2 + mg ( 2R )⎞ ,
⎜
⎟⎜
⎟
⎝
⎠⎝
⎠
1 or vf = Thus, v f = 1 2F ( π R )
+ v i2 + 4 gR .
m
2 ( 30.0 N ) π ( 0.600 m )
2
+ (15.0 m s ) + 4 9.80 m s 2 ( 0.600 m )
0.250 k g ( giving v f = 26.5 m s
142 ) CHAPTER 5.25 5 (a) When the rope is horizontal, the swing is 2.0 m above the level of the bottom of the
circular arc, and PEg = mgy = ( 40 N ) ( 2.0 m ) = 80 J .
(b) When the rope makes a 30° angle with the vertical, the vertical distance from the
swing to the lowest level in the circular arc is y = L − L cos 30° = L (1 − cos 30°) and the potential energy is given by PEg = mgy = mgL (1 − cos 30°) = ( 40 N ) ( 2.0 m ) (1 − cos 30°) = 11 J .
(c) At the bottom of the circular arc, y = 0 . Hence, PEg = 0 .
5.26 Using conservation of mechanical energy, we have
12
1
mv f + mgy f = m v i2 + 0 ,
2
2 or 5.27 yf = v i2 − v 2
f
2g = (10 m s ) − (1.0 m s )
2 ( 2 9.80 m s 2 2 = 5.1 m . ) Since no nonconservative forces do work, we use conservation of mechanical energy,
with the zero of potential energy selected at the level of the base of the hill. Then,
12
1
mv f + mgy f = m v i2 + m gy i with y f = 0 yields
2
2
yi = v 2 − v i2
f
2g = ( 3.00 ( m s) − 0
2 2 9.80 m s 2 ) = 0.459 m . Note that this result is independent of the mass of the child and sled.
5.28 (a) We take the zero of potential energy at the level of point B, and use conservation of
12
12
mechanical energy to obtain mv B + 0 = mv A + mgy A , or
2
2 ( ) 2
v B = v A + 2 gy A = 0 + 2 9.80 m s 2 ( 5.00 m ) = 9.90 m s . (b) At point C, with the starting point at A, we again use conservation of mechanical
12
12
energy. This gives mv C + m gy C = m v A + m gy A , and yields
2
2 ( ) 2
v C = v A + 2 g ( y A − y C ) = 0 + 2 9.80 m s 2 ( 3.00 m ) = 7.67 m s . 143 CHAPTER 5.29 5 Using conservation of mechanical energy, starting when point A is directly over the bar
and ending when it is directly below the bar, gives
12
1
mv f + mgy f = m v i2 + m gy i , or
2
2 ( ) ( ) v f = v i2 + 2 g y i − y f = 0 + 2 9.80 m s 2 ( 2.40 m ) = 6.86 m s 5.30 (a) From conservation of mechanical energy, A
B 12
12
mv B + mgy B = m v A + m gy A , or
2
2 5.00 m 3.20 m C
2.00 m v B = v + 2g ( y A − y B )
2
A ( ) = 0 + 2 9.80 m s 2 (1.80 m ) = 5.94 m s .
Similarly,
2
v C = v A + 2 g ( y A − y B ) = 0 + 2 g ( 5.00 m − 2.00 m ) = 7.67 m s . (b) W g 5.31 ) A →C ( ) − ( PE ) = PEg A g C = mg ( y A − y C ) = ( 49.0 N ) ( 3.00 m ) = 147 J (a) We choose the zero of potential energy at the level of the bottom of the arc. The
initial height of Tarzan above this level is y i = ( 30.0 m ) (1 − cos 37.0°) = 6.04 m .
Then, using conservation of mechanical energy, we find
12
1
mv f + 0 = m v i2 + mgy i , or
2
2 ( ) v f = v i2 + 2 gy i = 0 + 2 9.80 m s 2 ( 6.04 m ) = 10.9 m s .
(b) In this case, conservation of mechanical energy yields
v f = v i2 + 2 gy i = ( 4.00 ( ) m s ) + 2 9.80 m s 2 ( 6.04 m ) = 11.6 m s .
2 144 CHAPTER 5.32 5 Realize that all three masses have identical speeds at each point in the motion and that
v i = 0 . Then, conservation of mechanical energy gives KE f = PEi − PE f , or ( ( ) ) ( ) 1
( m1 + m2 + m3 ) v 2f = ⎡m1 y1i − y1 f + m2 y 2i − y 2 f + m3 y 3i − y 3 f ⎤ g
⎣
⎦
2 Thus, ( 1
( 30.0) v 2f = [ ( 5.00) ( −4.00 m ) + (10.0) ( 0) + (15.0) ( +4.00 m ) ] 9.80 m s 2
2 v f = 5.11 m s . yielding 5.33 ) (a) Use conservation of mechanical energy from when the projectile is at rest within the
gun until it reaches maximum height. ( Then, KE + PEg + PEs ) = ( KE + PE + PEs g f ) i becomes 1
0 + mgy max + 0 = 0 + 0 + kx i2 ,
2 or ( )( ) −3
2
2mgy max 2 20.0 × 10 kg 9.80 m s ( 20.0 m )
=
= 544 N m .
k=
x i2
( 0.120 m ) 2 (b) This time, we use conservation of mechanical energy from when the projectile is at
rest within the gun until it reaches the equilibrium position of the spring. This gives ( )( KE f = PEg + PEs − PEg + PEs
i ) f 1
⎛
⎞
= ⎜ −mgx i + kx i2 ⎟ − ( 0 + 0)
⎝
⎠
2 ⎛ k⎞
v 2 = ⎜ ⎟ x i2 − 2 gx i
f
⎝ m⎠
⎛ 544 N m ⎞
=⎜
( 0.120 m ) 2 − 2 9.80 m s 2 ( 0.120 m )
20.0 × 10−3 k g ⎟
⎝
⎠ ( yielding v f = 19.7 m s 145 ) CHAPTER 5.34 5 At maximum height, v y =0 and v x = v ix = ( 40 m s ) co s 60° = 20 m s .
2
2
Thus, v f = v x + v y = 20 m s . Choosing PEg = 0 at the level of the launch point, conservation of mechanical energy gives PE f = KEi − KE f , and the maximum height
reached is
yf = 5.35 v i2 − v 2
f
2g ( 40
= m s ) − ( 20 m s )
2 ( 2 9.80 m s 2 2 ) = 61 m . Choose PEg = 0 at the level of the release point and use conservation of mechanical
energy from release until the block reaches maximum height. Then, ( KE f = KEi = 0 and we have PEg + PEs ) = ( PE g f ) + PEs , or
i 1
mgy max + 0 = 0 + kx i2 which yields
2 y max 5.36 ( ) 5.00 × 103 N m ( 0.100 m )
kx i2
=
=
= 10.2 m .
2mg
2 ( 0.250 k g ) 9.80 m s 2 ( 2 ) (a) Choose PEg = 0 at the level of point B. Between A and B, we can use conservation of () mechanical energy, KEB + PEg B () = KEA + PEg A , which becomes 12
mv B + 0 = 0 + mgy A or
2 ( ) v B = 2 g y A = 2 9.80 m s 2 ( 5.00 m ) = 9.90 m s .
(b) Again, choose PEg = 0 at the level of point B. Between points B and C, we use the
workkinetic energy theorem in the form ( ) − ( KE + PE ) 12
= 0 + mgy C − m v B − 0 to find
2
1
2
2
Wnc = ( 0.400 k g ) 9.80 m s ( 2.00 m ) − ( 0.400 k g ) ( 9.90 m s ) = −11.8 J .
2
Wnc = KE + PEg g C ( B ) Thus, 11.8 J of energy is spent overcoming friction between B and C. 146 CHAPTER 5.37 5 The work done by the nonconservative resistance force is ( ) Wnc = ( f co s θ ) s = ⎡ 2 .7 × 10−3 N co s 180°⎤ ( 0.20 m ) = −5.4 × 10−4 J
⎣
⎦ We use the workkinetic theorem in the form ( Wnc = KE + PEg
Thus, g f i ( 12
1
mv f = Wnc + m v i2 + m g y i − y f
2
2
vf = = 5.38 ) − ( KE + PE ) , or KE 4.2 × 10 kg i f . ) )+ ( 2.5 × 10 2 −5.4 × 10−4 J
3 ( )( ) = Wnc + KEi + PEg − PEg ) or ( 2 Wnc
+ v i2 + 2 g y i − y f
m ( f −2 ) ( ) m s + 2 9.80 m s 2 ( 0.20 m ) = 1.9 m s
2 The friction force does work W friction on the gurney during the 20.7 m horizontal
displacement at constant speed. The attendant does 2000 J of work on the gurney during
this motion. Applying the workkinetic energy theorem gives
W nc = W frict ion + 2000 J = ( KE + PE) f − ( KE + PE)i = 0 , so W friction = −2000 J . Also, W frict ion = ( f co s 180° ) s = − f s . Therefore, the friction force is given by f= −W friction
s = − ( −2000 J)
= 96.6 N .
20.7 m Now consider the displacement s ′ = 0.48 m that occurs after the gurney is released with
an initial speed of v i = 0.88 m s . Applying the workkinetic energy theorem to this part
of the trip gives ( Wnc = ( f cos 180°) s′ = KE + PEg
or ) − ( KE + PE ) ,
g f i ⎛1
⎞
− f s′ = ( 0 + 0) − ⎜ mv i2 + 0⎟ which gives the total mass of the gurney plus patient
⎝2
⎠ as m = 2 f s′ v i2 , and the total weight as
⎡ 2 ( 96.6 N ) ( 0.48 m ) ⎤
⎛ 2 f s′ ⎞
⎥ 9.80 m s 2 = 1.2 × 103 N = 1.2 kN
w = mg = ⎜ 2 ⎟ g = ⎢
2
⎝ vi ⎠
⎢
( 0.88 m s ) ⎥
⎣
⎦ ( 147 ) CHAPTER 5.39 5 We shall take PEg = 0 at the lowest level reached by the diver under the water. The diver
falls a total of 15 m, but the nonconservative force due to water resistance acts only
during the last 5.0 m of fall. The workkinetic energy theorem then gives ( Wnc = KE + PEg
or ) − ( KE + PE ) ,
g f i ( F cos 180°) ( 5.0 m ) = ( 0 + 0) − ⎡0 + (70 kg ) ( 9.80 m s ) (15 m ) ⎤ .
⎣
⎦
2 This gives the average resistance force as F = 2.1 × 10 3 N = 2.1 k N . 5.40 ( ) = ( PE ) Since the plane is in level flight, PEg g f i and the workkinetic energy theorem reduces to Wnc = Wthrust + Wresistance = KE f − KEi , or ( F cos 0°) s + ( f cos 180) s = 2 mv 2f − 2 mv i2 .
1 1 This gives vf = v +
2
i 5.41 2( F − f ) s
m 2 ⎡( 7.5 − 4.0) × 10 4 N ⎤ ( 500 m )
⎦
= ( 60 m s ) + ⎣
= 77 m s
1.5 × 104 kg
2 Choose PEg = 0 at the level of the bottom of the driveway. ( Then Wnc = KE + PEg ) − ( KE + PE )
g f i becomes ⎡
⎤
( f cos 180°) s = ⎢ 2 mv 2f + 0⎥ − ⎡0 + mg ( s sin 20°) ⎤ .
⎣
⎦
1 ⎣ ⎦ Solving for the final speed gives v f = or ( v f = 2 9.80 m s 2 ( 2 gs) sin 20° − ) ( 5.0 m ) sin 20° − 148 ( 2f s
, or
m ) 2 4.0 × 103 N ( 5.0 m )
2.10 × 10 kg
3 = 3.8 m s . CHAPTER 5.42 5 (a) Choose PEg = 0 at the level of the bottom of the arc. The child’s initial vertical
displacement from this level is y i = ( 2.00 m ) (1 − cos 30.0°) = 0.268 m .
In the absence of friction, we use conservation of mechanical energy as ( KE + PE ) = ( KE + PE ) , or 1 mv
2
g g f i ( 2
f + 0 = 0 + mgy i , which gives ) v f = 2 gy i = 2 9.80 m s 2 ( 0.268 m ) = 2.29 m s .
(b) With a nonconservative force present, we use ( Wnc = KE + PEg ) − ( KE + PE ) = ⎛ 1 mv
⎜
⎝2
g f i 2
f ⎞
+ 0⎟ − ( 0 + mgy i ) , or
⎠ ⎛ v2
⎞
f
Wnc = m ⎜ − gy i ⎟
⎝2
⎠ ⎡ ( 2.00 m s ) 2
⎤
= ( 25.0 k g ) ⎢
− 9.80 m s 2 ( 0.268 m ) ⎥ = −15.6 J
2
⎢
⎥
⎣
⎦ ( ) Thus, 15.6 J of energy is spent overcoming friction. 5.43 ( (a) We use conservation of mechanical energy, KE + PEg ) = ( KE + PE ) , for the trip
g f down the frictionless ramp. With v i = 0 , this reduces to ( ) ( v f = 2 g y i − y f = 2 9.80 m s 2 ) [( 3.00 m ) sin 30.0°] = 5.42 m s . (b) Using the workkinetic energy theorem, ( Wnc = KE + PEg ) − ( KE + PE ) ,
g f i for the trip across the rough floor gives ( fk cos 180°) s = ( 0 + 0) − ⎛ 2 mvi2 + 0⎞ , or
⎜
⎟
⎝
⎠
1 149 1
fk s = µk ( m g ) s = m v i2 .
2 i CHAPTER 5 Thus, the coefficient of kinetic friction is ( 5.42 m s )
v2
µk = i =
= 0.300 .
2 g s 2 9.80 m s 2 ( 5.00 m )
2 ( ) ( ) (c) All of the initial mechanical energy, KE + PEg = 0 + mgy i , is spent overcoming
i friction on the rough floor. Therefore, the energy “lost” is ( ) mgy i = (10.0 m ) 9.80 m s 2 ⎡( 3.00 m ) sin 30.0°⎤ = 147 J .
⎣
⎦
5.44 Choose PEg = 0 at water level and ( use KE + PEg ) = ( KE + PE )
f g i for the trip down the curved slide.
This gives h θ
h/5 12
⎛ h⎞
mv + mg ⎜ ⎟ = 0 + mgh , so the
⎝ 5⎠
2
speed of the child as she leaves the end of the slide is v = y 2 g ( 4 h 5) . The vertical component of this launch velocity is
⎛ 4h ⎞
v iy = v sin θ = sin θ 2 g ⎜ ⎟ .
⎝ 5⎠
2
At the top of the arc, v y = 0 . Thus, v y = v i2y + 2ay ( ∆y ) gives the maximum height the child reaches during the airborne trip as h⎞
⎡ ⎛ 4h ⎞ ⎤
⎛
0 = sin 2 θ ⎢ 2 g ⎜ ⎟ ⎥ + 2 ( − g ) ⎜ y max − ⎟
⎝
5⎠
⎣ ⎝ 5 ⎠⎦
This may be solved for y max to yield y max = 150 ( ) h
4 s in 2 θ + 1 .
5 CHAPTER 5.45 5 Choose PEg = 0 at the level of the base of the hill and let x represent the distance the
skier moves along the horizontal portion before coming to rest. The normal force exerted
on the skier by the snow while on the hill is n1 = mg co s 10.5° and, while on the
horizontal portion, n 2 = mg .
Consider the entire trip, starting from rest at the top of the hill until the skier comes to
rest on the horizontal portion. The work done by friction forces is
Wnc = ⎡( f k )1 co s 180°⎤ ( 200 m ) + ⎡( f k ) 2 cos 180°⎤ x
⎣
⎦
⎣
⎦
= − µk ( mg co s 10.5°) ( 200 m ) − µk ( m g ) x ( Applying Wnc = KE + PEg ) − ( KE + PE )
g f i to this complete trip gives − µk ( mg co s 10.5°) ( 200 m ) − µk ( mg ) x = [ 0 + 0] − ⎡0 + m g ( 200 m ) sin 10.5°⎤ ,
⎣
⎦
⎛ sin 10.5°
⎞
− co s 10.5°⎟ ( 200 m ) . If µ k = 0.0750 , then x = 289 m .
x =⎜
⎝ µk
⎠ or 5.46 The normal force exerted on the sled by the track is n = m g co s θ and the friction force is fk = µk n = µk mg cos θ .
If s is the distance measured along the incline that the sled travels, applying ( Wnc = KE + PEg ) − ( KE + PE )
f g i to the entire trip gives ⎡1 2
⎤
⎡
⎤⎣
⎦
⎣( µk mg cos θ ) cos 180°⎦ s = ⎡0 + mg s ( sin θ ) ⎤ − ⎢ 2 mv i + 0⎥ ,
⎣
⎦ ( 4.0 m s )
v i2
=
= 1.5 m .
2
2 g ( sin θ + µk co s θ ) 2 9.80 m s ( sin 20° + 0.20 cos 20°)
2 or s= ( ) 151 CHAPTER 5.47 5 ( (a) Consider the entire trip and apply Wnc = KE + PEg ) − ( KE + PE )
g f i to obtain ( f1 cos 180°) d1 + ( f2 cos 180°) d2 = ⎛ 2 mv 2f + 0⎞ − ( 0 + mgyi ) , or
⎜
⎟
⎝
⎠
1 vf = f d + f2 d2 ⎞
⎛
2⎜ g yi − 1 1
⎟
⎝
⎠
m ⎛
( 50.0 N ) ( 800 m ) + ( 3600 N ) ( 200 m ) ⎞
= 2 ⎜ 9.80 m s 2 (1000 m ) −
⎟
80.0 k g
⎝
⎠ ( ) which yields v f = 24.5 m s .
(b) Yes , this is too fast for safety. ( (c) Again, apply Wnc = KE + PEg ) − ( KE + PE ) , now with d
g f 2 i considered to be a variable, d1 = 1000 m − d2 , and v f = 5.00 m s . This gives ( f1 cos 180°) (1000 m − d2 ) + ( f2 cos 180°) d2 = ⎛ 2 mv 2f + 0⎞ − ( 0 + mgyi ) ,
⎜
⎟
⎝
⎠
1 1
which reduces to − (1000 m ) f1 + f1 d2 − f2 d2 = m v 2 − mgy i . Therefore,
f
2 d2 = = ( mg) yi − (1000 m ) f1 − 1 mv 2f
2
f2 − f1 ( 784 N ) (1000 m ) − (1000 m ) ( 50.0 N ) − ( 80.0 k g ) ( 5.00 m s )
3600 N − 50.0 N 1
2 2 = 206 m (d) In reality. the air drag will depend on the skydiver’s speed. It will be larger than her
784 N weight only after the chute is opened. It will be nearly equal to 784 N before
she opens the chute and again before she touches down, whenever she moves near
terminal speed.
5.48 (a) W nc = ∆KE + ∆PE , but ∆KE = 0 because the speed is constant. The skier rises a
vertical distance of ∆y = ( 60 m ) sin 30° = 30 m . Thus, ( ) Wnc = ( 70 kg ) 9.80 m s 2 ( 30 m ) = 2 .06 × 104 J = 21 kJ .
152 CHAPTER 5 (b) The time to travel 60 m at a constant speed of 2 .0 m s is 30 s. Thus, the required
power input is ℘= 5.49 Wnc 2.06 × 104 J
⎛ 1 hp ⎞
=
= ( 686 W ) ⎜
= 0.92 hp .
⎝ 746 W ⎟
⎠
∆t
30 s (a) The work done by the student is ( ) W = ∆PEg = mg ( ∆y ) = ( 50.0 kg ) 9.80 m s 2 ( 5.00 m ) = 2 .45 × 103 J .
The time to do this work if she is to match the power output of a 200 W light bulb is
W 2 .45 × 103 J
t=
=
= 12 .3 s , so the required average speed is
℘
200 W
v= ∆y 5.00 m
=
= 0.408 m s .
12 .3 s
t (b) The work done was found in part (a) as W = 2 .45 × 103 J= 2.45 kJ . 5.50 Let ∆N be the number of steps taken in time ∆t. We determine the number of steps per
unit time by
Po w er = or wo r k d o n e ( w o r k p e r st ep p e r u n it m a ss ) ( ma ss ) ( # st ep s )
,
=
∆t
∆t ⎛
J st ep ⎞
⎛ ∆N ⎞
70 W = ⎜ 0.60
⎟ ( 60 kg ) ⎜ ∆t ⎟ , giving
⎝
⎠
kg ⎠
⎝ ∆N
= 1.9 st ep s s .
∆t The running speed is then v= 5.51 st ep ⎞ ⎛
∆x ⎛ ∆N ⎞
m⎞
⎛
=⎜
= 2.9 m s
⎟ ( d ist a n ce t r a v eled p er st ep ) = ⎜ 1.9
⎟ ⎜ 1.5
⎝
∆t ⎝ ∆t ⎠
s ⎠⎝
st ep ⎟
⎠ The power output is given by ℘=
or energy spen t ( ∆m ) g h
=
= 1.2 × 106 k g s 9.80 m s 2 ( 50 m ) ,
∆t
∆t ( )( ℘= 5.9 × 108 W= 5.9 × 102 MW . 153 ) CHAPTER 5.52 5 (a) We use the workkinetic energy theorem with KEi = 0 a n d ∆PE = 0 to find 1
2
⎛1
⎞
W = ∆KE + ∆PE = ⎜ mv 2 − 0⎟ + ( 0) = (1500 kg ) (10.0 m s )
⎝2
⎠
2 W = 7.50 × 104 J = 75.0 kJ .
(b) The average power input is given by
℘= W 7.50 × 104
=
∆t
3.00 s J ⎛ 1 hp ⎞
= 33.5 h p .
⎜
⎝ 746 W ⎟
⎠ (c) The acceleration of the car is
a= ∆v 10.0 m s − 0
=
= 3.33 m s 2 .
∆t
3.00 s Thus, the net force acting on the car is ( )( ) F = ma = 1.50 × 103 kg 3.33 m s 2 = 5.00 × 103 N .
At t = 2.00 s, the instantaneous velocity is ( ) v = v i + at = 0 + 3.33 m s 2 ( 2.00 s ) = 6.66 m s
and the instantaneous power input is
⎛ 1 hp ⎞
℘ = Fv = 5.00 × 103 N ( 6.66 m s ) ⎜
= 44.7 hp
⎝ 746 W ⎟
⎠ ( 5.53 ) (a) The acceleration of the car is a = v f − vi
t = 18.0 m s − 0
= 1.50 m s 2 . Thus, the
12 .0 s constant forward force due to the engine is found from ( . )( ΣF = Fengine − Fair = ma as ) Fengine = Fair + ma = 400 N + 1.50 × 103 k g 1.50 m s 2 = 2 .65 × 103 N
The average velocity of the car during this interval is v = v f + vi
2 average power input from the engine during this time is
⎛ 1 hp ⎞
℘= Fengine v = 2 .65 × 103 N ( 9.00 m s ) ⎜
= 32.0 hp .
⎝ 746 W ⎟
⎠ ( ) 154 = 9.00 m s , so the CHAPTER 5 (b) At t = 12 .0 s , the instantaneous velocity of the car is v = 18.0 m s and the
instantaneous power input from the engine is
⎛ 1 hp ⎞
℘= Fengine v = 2 .65 × 103 N (18.0 m s ) ⎜
= 63.9 hp .
⎝ 746 W ⎟
⎠ ( 5.54 ) (a) The acceleration of the elevator during the first 3.00 s is a= v f − vi
t = 1.75 m s − 0
= 0.583 m s 2 ,
3.00 s so Fnet = Fmot or − mg = ma gives the force exerted by the motor as Fmotor = m ( a + g ) = ( 650 kg ) ⎡( 0.583 + 9.80) m s 2 ⎤ = 6.75 × 103 N .
⎣
⎦
The average velocity of the elevator during this interval is v = v f + vi
2 = 0.875 m s , so the average power input from the motor during this time is
⎛ 1 hp ⎞
℘= Fmot or v = 6.75 × 103 N ( 0.875 m s ) ⎜
= 7.92 hp .
⎝ 746 W ⎟
⎠ ( ) (b) When the elevator moves upward with a constant speed of v = 1.75 m s , the
upward force exerted by the motor is Fmot or = mg and the instantaneous power input
from the motor is
⎛ 1 hp ⎞
℘= ( mg ) v = ( 650 k g ) 9.80 m s 2 (1.75 m s ) ⎜
= 14.9 hp .
⎝ 746 W ⎟
⎠ ( 5.55 ) The work done on the particle by the force F as
the particle moves from x = x i to x = x f is the area
under the curve from x i to x f . Fx (N) 4
2 (a) For x = 0 to x = 8.00 m , 0 1
W = ar ea o f t r ia n g le A BC = A C × a ltitu d e
2
W0→ 8 = B 6 1
( 8.00 m ) ( 6.00 N ) = 24.0 J
2 155 –2
–4 A C
2 4 6 E 8 10 12
D x (m) CHAPTER 5 (b) For x = 8.00 m to x = 10.0 m ,
1
W8→10 = ar ea o f t r ia n g le CDE = CE × a ltitu d e
2
= 1
( 2 .00 m ) ( −3.00 N ) = − 3.00 J
2 (c) W0→10 = W0→8 + W8→10 = 24.0 J + ( − 3.00 J) = 21.0 J
5.56 W equals the area under the ForceDisplacement Curve
(a) For the region 0 ≤ x ≤ 5.00 m , W= ( 3.00 N ) ( 5.00 m )
2 Fx (N)
3 = 7.50 J 2
1 (b) For the region 5.00 m ≤ x ≤ 10.0 m , 0 W = ( 3.00 N ) ( 5.00 m ) = 15.0 J (c) For the region 10.0 m ≤ x ≤ 15.0 m , W = x (m)
2 0 4 6 ( 3.00 N ) ( 5.00 m )
2 8 10 12 14 16 = 7.50 J (d) For the region 0 ≤ x ≤ 15.0 m , W = ( 7.50 + 15.0 + 7.50) J = 30.0 J
5.57 (a) Fx = ( 8 x − 16) N (b) Wnet = Fx (N) − ( 2.00 m ) (16.0 N )
2 + (1.00 m ) ( 8.00 N )
2 20 = − 12 .0 J 0 5.58 (3, 8) 10 –10 1 From the workkinetic energy theorem, Wnet = KE f − KEi , we have
–20 (F applied ) 1
co s 0° s = m v 2 − 0 .
f
2 The mass of the cart is m = vf = 2 Fapplied s m = w
= 10.0 kg , and the final speed of the cart is
g 2 ( 40.0 N ) (12 .0 m )
10.0 k g 156 = 9.80 m s . 2 3 4 x (m) CHAPTER 5.59 5 (a) The equivalent spring constant of the bow is given by F = kx as
k= Ff = xf 230 N
= 575 N m .
0.400 m (b) From the workkinetic energy theorem applied to this situation, ( Wnc = KE + PEg + PEs ) − ( KE + PE g f + PEs ) f 1
⎛
⎞
= ⎜ 0 + 0 + kx 2 ⎟ − ( 0 + 0 + 0) .
f
⎝
⎠
2 The work done pulling the bow is then
1
1
2
Wnc = kx 2 = ( 575 N m ) ( 0.400 m ) = 46.0 J .
f
2
2 5.60 Choose PEg = 0 at the level where the block comes to rest against the spring. Then, in
the absence of work done by nonconservative forces, the conservation of mechanical
energy gives ( KE + PE g or + PEs ) = ( KE + PE g f ) + PEs ,
i 1
0 + 0 + kx 2 = 0 + mg L sin θ + 0 . Thus,
f
2 ( ) 2 (12.0 k g ) 9.80 m s 2 ( 3.00 m ) sin 35.0°
2 mg L sin θ
=
= 0.116 m .
xf =
3.00 × 104 N m
k 5.61 (a) From v 2 = v i2 + 2ay ( ∆y ) , we find the speed just before touching the ground as ( ) v = 0 + 2 9.80 m s 2 (1.0 m ) = 4.4 m s .
(b) Choose PEg = 0 at the level where the feet come to rest. Then ( Wnc = KE + PEg ) − ( KE + PE )
g f i ( F cos 180°) s = ( 0 + 0) − ⎛ 1 mv
⎜
⎝2 becomes 2
i ⎞
+ mg s⎟
⎠ (75 kg ) ( 4.4 m s ) + 75 kg 9.80 m s 2 = 1.5 × 105 N
mv i2
or F =
+ mg =
(
)
2s
2 5.0 × 10−3 m
2 ( ) 157 ( ) CHAPTER 5.62 5 From the workkinetic energy theorem, ( Wnc = KE + PEg + PEs ) − ( KE + PE g f ) + PEs ,
i 1
⎛1
⎞⎛
⎞
we have ( fk cos 180°) s = ⎜ mv 2 + 0 + 0⎟ − ⎜ 0 + 0 + kx i2 ⎟ , or
f
⎝2
⎠⎝
⎠
2 vf = 5.63 ( 8.0 kx i2 − 2 f k s
=
m ( N m ) 5.0 × 10 −2 m ) 2 5.3 × 10 − 2 ( 0.032 N ) ( 0.15 m ) −3 kg = 1.4 m s (a) The two masses will pass when both are at y f = 2.00 m above the floor. From ( conservation of energy, KE + PEg + PEs ) = ( KE + PE g f ) + PEs ,
i 1
( m1 + m2 ) v 2f + ( m1 + m2 ) gy f + 0 = 0 + m1 gy 1i + 0 , or
2
vf = = 2 m1 g y 1i
− 2gyf
m1 + m 2 ( ) 2 ( 5.00 k g ) 9.80 m s 2 ( 4.00 m )
8.00 k g ( ) − 2 9.80 m s 2 ( 2 .00 m ) . This yields the passing speed as v f = 3.13 m s .
(b) When m1 = 5.00 k g reaches the floor, m2 = 3.00 kg is y 2 f = 4.00 m above the floor. ( Thus, KE + PEg + PEs ) = ( KE + PE
f g + PEs ) i becomes 1
( m1 + m2 ) v 2f + m2 g y 2 f + 0 = 0 + m1 gy 1i + 0 , or v f =
2 ( 2 g m 1 y 1i − m 2 y 2 f ) m1 + m 2 Thus, vf = ( ) 2 9.80 m s 2 ⎡( 5.00 kg ) ( 4.00 m ) − ( 3.00 k g ) ( 4.00 m ) ⎤
⎣
⎦ = 4.43 m s
8.00 kg 158 CHAPTER 5 (c) When the 5.00kg hits the floor, the string goes slack and the 3.00kg becomes a
projectile launched straight upward with initial speed v iy = 4.43 m s . At the top of
2
its arc, v y = v i2y + 2ay ( ∆y ) gives ∆y = 5.64 2
v y − v i2y 2 ay = 0 − ( 4.43 m s ) ( 2 −9.80 m s 2 2 = 1.00 m . ) The normal force the incline exerts on block A is n A = ( m A g ) co s 37° , and the friction force
is fk = µk n A = µk mA g cos 37° . The vertical distance block A rises is ∆y A = ( 20 m ) sin 37° = 12 m , while the vertical displacement of block B is ∆y B = −20 m . We find the common final speed of the two blocks by use of ( Wnc = KE + PEg ) − ( KE + PE ) = ∆KE + ∆PE .
g f g i ⎡1
⎤
This gives − ( µk mA g cos 37°) s = ⎢ ( mA + mB ) v 2 − 0⎥ + ⎡mA g ( ∆y A ) + mB g ( ∆y B ) ⎤
f
⎦
⎣2
⎦⎣ or v2 =
f 2 g ⎡−mB ( ∆y B ) − mA ( ∆y A ) − ( µk mA cos 37°) s ⎤
⎣
⎦
m A + mB ( ) 2 9.80 m s 2 ⎡− (100 kg ) ( −20 m ) − ( 50 kg ) (12 m ) − 0.25 ( 50 kg ) ( 20 m ) cos 37°⎤
⎣
⎦
=
150 kg
which yields v 2 = 157 m 2 s 2 .
f
The change in the kinetic energy of block A is then ( ) 1
1
∆KEA = m A v 2 − 0 = ( 50 k g ) 157 m 2 s 2 = 3.9 × 103 J = 3.9 kJ .
f
2
2 5.65 Since the Marine moves at constant speed, the upward force the rope exerts on him must
equal his weight, or F = 700 N . The constant speed at which he moves up the rope is
∆y 10.0 m
v=
=
= 1.25 m s , so the constant power output is
∆t
8.00 s
℘ = Fv = ( 700 N ) (1.25 m s ) = 875 W . 159 CHAPTER 5.66 5 When 1 pound (454 grams) of fat is metabolized, the energy released is
E = ( 454 g ) ( 9.00 k ca l g ) = 4.09 × 10 3 k ca l . Of this, 20.0% goes into mechanical energy
(climbing stairs in this case). Thus, the mechanical energy generated by metabolizing 1
pound of fat is ( ) Em = ( 0.200) 4.09 × 103 kca l = 817 kca l .
Each time the student climbs the stairs, she raises her body a vertical distance of
∆y = ( 80 st e p s ) ( 0.150 m st ep ) = 12.0 m . The mechanical energy required to do this is
∆PEg = m g ( ∆y ) , or ⎛ 1 kca l ⎞
∆PEg = ( 50.0 kg ) 9.80 m s 2 (12.0 m ) = 5.88 × 103 J ⎜
= 1.40 kcal .
⎝ 4186 J⎟
⎠ ( ( ) ) (a) The number of times the student must climb the stairs to metabolize 1 pound of fat
E
817 kca l
= 582 t r ip s .
is N = m =
∆PEg 1.40 kca l tr ip
It would be more practical for her to reduce food intake.
(b) The useful work done each time the student climbs the stairs is W = ∆PEg = 5.88 × 10 3 J .
Since this is accomplished in 65.0 s, the average power output is ℘= 5.67 W 5.88 × 103 J
⎛ 1 hp ⎞
=
= 90.5 W = ( 90.5 W ) ⎜
= 0.121 hp .
⎝ 746 W ⎟
⎠
t
65.0 s ⎛ 4186 J⎞
= 9.21 × 105 J of energy while going
(a) The person walking uses Ew = ( 220 kca l ) ⎜
⎝ 1 kca l ⎟
⎠
3.00 miles. The quantity of gasoline which could furnish this much energy is
9.21 × 105 J
V1 =
= 7.08 × 10−3 g a l . This means that the walker’s fuel economy in
1.30 × 108 J g a l
equivalent miles per gallon is
fu el econ om y = 3.00 m i
= 423 m i g a l .
7.08 × 103 g a l 160 CHAPTER 5 (b) In 1 hour, the bicyclist travels 10.0 miles and uses ⎛ 4186 J⎞
EB = ( 400 kca l ) ⎜
= 1.67 × 106 J ,
⎝ 1 kca l ⎟
⎠
which is equal to the energy available in 1.67 × 106 J
= 1.29 × 10−2 g a l of gasoline. Thus, the equivalent fuel economy for
1.30 × 108 J g a l
10.0 m i
the bicyclist is
= 776 m i g a l .
1.29 × 10−2 g a l
V2 = 5.68 We use Wnet = KE f − KEi to find the average force the hand exerts on the baseball. m v2
1
Wnet = F co s 180° s = 0 − mball v i2 , so F = ball i .
2s
2 ( ) The force the ball exerts on the hand is in the opposite direction but has the same
magnitude.
(a) If s = 2.0 cm = 2.0 × 10−2 m , ( 0.15 kg ) ( 25 m s )
F= ( 2 2 .0 × 10 −2 m 2 = 2.3 × 103 N = 2.3 kN . ) 2
(b) Similarly, if s = 10 cm = 0.10 m , we find F = 4.7 × 10 N . 5.69 ( (a) Use conservation of mechanical energy, KE + PEg ) = ( KE + PE ) , from the start to
f g i the end of the track to find the speed of the skier as he leaves the track. This gives
12
mv + mgy f = 0 + mgy i , or
2
v= ( ) 2 g yi − y f = ( ) 2 9.80 m s 2 ( 40.0 m ) = 28.0 m s . 161 CHAPTER 5 (b) At the top of the parabolic arc the skier follows after leaving the track, v y = 0 and
2
2
v x = ( 28.0 m s ) co s 45.0° = 19.8 m s . Thus, v top = v x + v y = 19.8 m s . Applying conservation of mechanical energy from the end of the track to the top of the arc
1
1
2
2
gives m (19.8 m s ) + m g y max = m ( 28.0 m s ) + m g (10.0 m ) , or
2
2 y max ( 28.0
= 10.0 m + m s ) − (19.8 m s )
2 ( 2 9.80 m s 2 ) 2 = 30.0 m . 1
(c) Using ∆y = v iy t + ay t 2 for the flight from the end of the track to the ground gives
2 ( ) 1
−10.0 m = ⎡( 28.0 m s ) sin 45.0°⎤ t + −9.80 m s 2 t 2
⎣
⎦2 The positive solution of this equation gives the total time of flight as t = 4.49 s .
During this time, the skier has a horizontal displacement of ∆x = v ix t = ⎡( 28.0 m s ) cos 45.0°⎤ ( 4.49 s ) = 89.0 m .
⎣
⎦
5.70 First, determine the magnitude of the applied force by considering
a freebody diagram of the block. Since the block moves with
constant velocity, ΣFx = Fy = 0 .
From ΣFx = 0 , we see that n = F co s 30° . n
30° 5 kg
fk F
mg Thus, fk = µk n = µk F cos 30° , and ΣFy = 0 becomes
F sin 30° = m g + µk F co s 30° , or F= ( ) ( 5.0 kg ) 9.80 m s 2
mg
=
= 2.0 × 102 N .
sin 30° − µk cos 30° sin 30° − ( 0.30) cos 30° (a) The applied force makes a 60° angle with the displacement up the wall. Therefore, ( ) 2
W F = ( F co s 60°) s = ⎡ 2 .0 × 10 2 N co s 60°⎤ ( 3.0 m ) = 3.1 × 10 J .
⎣
⎦
2
(b) W g = ( m g co s 180°) s = ( 49 N ) ( −1.0) ( 3.0 m ) = − 1.5 × 10 J . (c) Wn = ( n co s 90°) s = 0 .
162 CHAPTER 5 2
(d) PEg = m g ( ∆y ) = ( 49 N ) ( 3.0 m ) = 1.5 × 10 J . 5.71 The force constant of the spring is k = 1.20 N cm = 120 N m . If the spring is initially
compressed a distance x i , the vertical distance the ball rises as the spring returns to the
equilibrium position is y f = x i sin 10.0° ( In the absence of friction, we apply KE + PEg + PEs ) = ( KE + PE
f g + PEs ) i from the release of the plunger to when the spring has returned to the equilibrium position and obtain
12
1
mv f + mg ( x i sin 10.0°) + 0 = 0 + 0 + kx i2 , or
2
2 vf = = This yields 5.72 kx i2
− 2 g x i sin 10.0°
m (120 ( N m ) 5.00 × 10 −2 m
0.100 kg ) 2 ( )( ) − 2 9.80 m s 2 5.00 × 10 −2 m sin 10.0° v f = 1.68 m s . If a projectile is launched, in the absence of air resistance, with speed v i at angle θ
above the horizontal, the time required to return to the original level is found from
g
2 v s in θ
1
. The range is the
∆y = v iy t + ay t 2 as 0 = ( v i sin θ ) t − t 2 , which gives t = i
2
2
g
horizontal displacement occurring in this time. ⎛ 2 v sin θ ⎞ v i2 ( 2 sin θ co s θ ) v i2 sin ( 2θ )
=
=
.
Thus, R = v ix t = ( v i co s θ ) ⎜ i
g⎟
g
g
⎝
⎠
Maximum range occurs when θ = 45° , giving vi2 = g Rmax . The minimum kinetic energy
required to reach a given maximum range is
1
1
KE = mv i2 = mg Rmax
2
2 163 CHAPTER 5 (a) The minimum kinetic energy needed in the record throw of each object is
KE = Javelin: ( ) 1
( 0.80 kg ) 9.80 m s 2 ( 89 m ) = 3.5 × 102 J
2 ( ) ( ) Discus: KE = 1
( 2.0 kg ) 9.80 m s 2 ( 69 m ) = 6.8 × 102 J
2 KE = 1
(7.2 kg ) 9.80 m s 2 ( 21 m ) = 7.4 × 102 J
2 Shot: (b) The average force exerted on an object during launch, when it starts from rest and is
given the kinetic energy found above, is computed from Wnet = F s = ∆KE as
F= KE − 0
. Thus, the required force for each object is
s F= Javelin: 3.5 × 102 J
2
= 1.7 × 10 N
2.00 m Discus: F = 6.8 × 102 J
2
= 3.4 × 10 N
2.00 m F= 7.4 × 102 J
2
= 3.7 × 10 N
2.00 m Shot:
(c) Yes . If the muscles are capable of exerting 3.7 × 10 2 N on an object and giving that
object a kinetic energy of 7.4 × 102 J , as in the case of the shot, those same muscles
should be able to give the javelin a launch speed of vi = 2 KE
=
m ( ) = 43 m 2 7.4 × 102 J
0.80 kg s, v 2 ( 43 m s )
= i=
= 1.9 × 102 m .
2
g 9.80 m s
2 with a resulting range of Rmax Since this far exceeds the record range for the javelin, one must conclude that air
resistance plays a very significant role in these events. 164 CHAPTER 5.73 The potential energy associated with the
wind is PEw = F x , where x is measured
horizontally from directly below the pivot
of the swing and positive when moving into
the wind, negative when moving with the
wind. We choose PEg = 0 at the level of the 5 wind direction L cosφ L cosθ L θ φ pivot as shown in the figure. Also, note that
D = L sin φ + L s in θ vi
x=0 ⎛D
⎞
so φ = sin −1 ⎜ − sin θ ⎟ , or
⎝L
⎠ L sinφ L sinθ
D 50.0 m
⎞
φ = s in ⎛
− sin 50.0°⎟ = 28.94° .
⎜
⎝ 40.0 m
⎠
−1 (a) Use conservation of mechanical energy, including the potential energy associated
with the wind. The final kinetic energy is zero if Jane barely makes it to the other ( side, and KE + PEg + PEw ) = ( KE + PE + PEw g f ) i becomes 1
0 + mg ( − L cos φ ) + F ( + L sin φ ) = mv i2 + mg ( − L cos θ ) + F ( − L sin θ ) ,
2 or v i = 2 gL ( co s θ − co s φ ) + 2FL
( s i n θ + s in φ )
m where m is the mass of Jane alone. This yields v i = 6.15 m s .
(b) Again, using conservation of mechanical energy with KE f = 0 , ( KE + PE g + PEw ) = ( KE + PE
f g + PEw 0 + Mg ( − L cos θ ) + F ( − L sin θ ) = ) i gives 1
M v i2 + M g ( − L cos φ ) + F ( + L sin φ )
2 where M = 130 k g is the combined mass of Tarzan and Jane. Thus,
vi = 2 gL ( co s φ − co s θ ) − 2FL
( sin θ + sin φ ) which gives v i = 9.87 m s
M 165 CHAPTER 5.74 5 The power loss is the rate at which work is done moving air, or ℘loss 1
∆m v 2
W ∆KE 2 ( )
. But the mass of air moved in time ∆t is
=
=
=
∆t
∆t
∆t ∆m = ( den sit y ⋅ v olu m e) = ρ ( area ⋅ len gt h ) = ρ A ∆x = ρ A v ( ∆t ) . Thus, ( ∆m) v 2 [ ρA v ( ∆t ) ] v 2
℘loss =
=
=
2 ( ∆t )
2 ( ∆t ) 1
ρA v 3 .
2 When an object is moved at constant speed, the power expenditure is ℘loss = fresistance v , so
the resistance force is given by
fresist ance = 5.75 3
1
℘loss (1 2) ρ Av
=
= ρA v 2 .
2
v
v We choose PEg = 0 at the level where the spring is relaxed (x = 0), or at the level of
position B.
(a) At position A, KE = 0 and the total energy of the system is given by ( E = 0 + PEg + PEs ) 12
= m g x 1 + k x 1 , or
2 A ( ) E = ( 25.0 k g ) 9.80 m s 2 ( −0.100 m ) + ( ) 1
2
2.50 × 10 4 N m ( −0.100 m ) = 101 J
2 (b) In position C, KE = 0 and the spring is uncompressed so PEs = 0 . ( Hence, E = 0 + PEg + 0
or x 2 = ) C = mg x 2 E
101 J
=
= 0.410 m .
mg ( 25.0 kg ) 9.80 m s 2 ( ) 12
(c) At Position B, PEg = PEs = 0 and E = ( KE + 0 + 0) B = m v B .
2 Therefore, v B = 2E
=
m 2 (101 J)
= 2.84 m s .
25.0 k g 166 CHAPTER 5 (d) Where the velocity (and hence the kinetic energy) is a maximum, the acceleration is
ΣFy
ay =
= 0 (at this point, an upward force due to the spring exactly balances the
m
downward force of gravity). Thus, taking upward as positive, ΣFy = −kx − mg = 0 or
x=− 245 k g
mg
=−
= −9.80 × 10−3 m = –9.80 mm .
2.50 × 10 4 N m
k 1
1
(e) From the total energy, E = KE + PEg + PEs = m v 2 + m g x + k x 2 , we find
2
2
v= 2E
k
− 2 g x − x2
m
m Where the speed, and hence kinetic energy is a maximum (i.e., at x = − 9.80 m m ),
this gives v m a x = 2.85 m s
5.76 When the block moves distance x down the incline, the work done by the friction force is
W f = ( f k co s 180°) x = − µk n x = − µk ( m g co s θ ) x . From the workkinetic energy theorem, ( Wnc = KE + PEg + PEs ) − ( KE + PE
f g ) + PEs , we find
i Wnc = W f = − µk ( m g co s θ ) x = ∆KE + ∆PEg + ∆PEs . Since the block is at rest at both the start and the end, this gives − µk (19.6 N cos 37.0°) ( 0.200 m )
= 0 + (19.6 N ) ( −0.200 m sin 37.0°) +
or
5.77 1
(100 N m ) ( 0.200 m ) 2
2 µ k = 0.115 . Choose PEg = 0 at the level of the river. Then y i = 36.0 m , y f = 4.00 m , the jumper falls
32.0 m, and the cord stretches 7.00 m. Between the balloon and the level where the diver ( stops momentarily, KE + PEg + PEs ) = ( KE + PE g f + PEs ) i gives 1
2
0 + ( 700 N ) ( 4.00 m ) + k ( 7.00 m ) = 0 + ( 700 N ) ( 36.0 m ) + 0 ,
2 or k = 914 N m . 167 CHAPTER 5.78 5 (a) Since the tension in the string is always perpendicular to the motion of the object,
the string does no work on the object. Then, mechanical energy is conserved: ( KE + PE ) = ( KE + PE ) .
g g f i Choosing PEg = 0 at the level where the string attaches to the cart, this gives
12
0 + mg ( − L co s θ ) = mv o + mg ( − L) , or v 0 =
2 2 g L ( 1 − co s θ ) (b) If L = 1.20 m and θ = 35.0° , the result of part (a) gives v0 = ( ) 2 9.80 m s 2 (1.20 m ) (1 − cos 35.0°) = 2.06 m s . 168 CHAPTER 5 Answers to Even Numbered Conceptual Questions
2. (a) The chicken does positive work on the ground. (b) No work is done. (c) The crane does
positive work on the bucket. (d) The force of gravity does negative work on the bucket.
(e) The leg muscles do negative work on the individual. 4. (a) Kinetic energy is always positive. Mass and speed squared are both positive.
(b) Gravitational potential energy can be negative when the object is below (i.e., closer to
the Earth) the chosen zero level. 6. (a) Kinetic energy is proportional to the speed squared. Doubling the speed makes the
object’s kinetic energy four times larger. (b) If the total work done on an object in some
process is zero, its speed must be the same at the final point as it was at the initial point. 8. The total energy of the bowling ball is conserved. Because the ball initially has
gravitational potential energy mgh and no kinetic energy, it will again have zero kinetic
energy when it returns to its original position. Air resistance and friction at the support
will cause the ball to come back to a point slightly below its initial position. On the other
hand, if anyone gives a forward push to the ball anywhere along its path, the
demonstrator will have to duck. 10. The effects are the same except for such features as having to overcome air resistance
outside. The person must lift his body slightly with each step on the tilted treadmill. Thus,
the effect is that of running uphill. 12. Both the force of kinetic friction exerted on the sled by the snow and the resistance force
exerted on the moving sled by the air will do negative work on the sled. Since the sled is
maintaining constant velocity, some towing agent must do an equal amount of positive
work, so the net work done on the sled is zero. 14. The kinetic energy is converted to internal energy within the brake pads of the car, the
roadway, and the tires. 16. The kinetic energy is a maximum at the instant the ball is released. The gravitational
potential energy is a maximum at the top of the flight path. 18. The normal force is always perpendicular to the surface and the motion is generally
parallel to the surface. Thus, in most circumstances, the normal force is perpendicular to
the displacement and does no work. The force of static friction does no work because there
is no displacement of the object relative to the surface in a static situation. 169 CHAPTER 5 Answers to Even Numbered Problems
2. 30.6 m 4. 1.6 kJ 6. 0.68 J 8. (a) 31.9 J (b) 0 (c) 10. (a) 14. (a) 18. (a) 22. 26.5 m s 26. 5.1 m 28. (a) 9.90 m s 30. (a) v B = 5.94 m s , v C = 7.67 m s 32. 5.11 m s 34. 61 m 36. (a) 38. 1.2 kN 40. 77 m s 42. (a) 2 .29 m s 44. h
4 s in 2 θ + 1
5 ( ) (e) 58.6 J 24. 148 J 2.0 m 20. (d) 90.0 J 16. 31.9 J 160 m s 12. 0 (d) –168 J (b) –184 J (c) 500 J (b) 19.6 J 9.90 m s 5.0 m s (c) 6.3 J (b) 1.2 J 39.2 J (c) 0 (b) 7.67 m s (b) 147 J 11.8 J (b) (b) 15.6 J 170 5.64 m s CHAPTER 46. 1.5 m (measured along the incline) 48. (a) 50. 2 .9 m s 52. (a) 54. 5 21 kJ (b) 0.92 hp 75.0 kJ (b) 33.5 hp (a) 7.92 hp (b) 14.9 hp 56. (a) 7.50 J (b) 15.0 J 58. 9.80 m s 60. 0.116 m 62. 1.4 m s 64. 3.9 kJ 66. (a) 582 trips (b) 90.5 W (0.121 hp) 68. (a) 2.3 kN (b) 4.7 × 10 2 N 70. (a) 3.1 × 102 J (b) −1.5 × 102 J 72. (a) 3.5 × 102 J for javelin, 6.8 × 102 J for discus, 7.4 × 102 J for shot
(b) 1.7 × 10 2 N on javelin, 3.4 × 10 2 N on discus, 3.7 × 10 2 N on shot
(c) Yes 76. 0.115 78. (b) (c) 44.7 hp (c) 7.50 J (c) 2 .06 m s 171 0 (d) (d) 30.0 J 1.5 × 102 J 172 ...
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 Spring '05
 Dr.Ha
 Physics, Force, Work

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