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Energy of a System
CHAPTER OUTLINE 7.2 Work Done by a Constant Force 7.3 The Scalar Product of Two Vectors 7.4 Work Done by a Varying Force 7.5 Kinetic Energy and the Work- Kinetic Energy Theorem 7.6 Potential Energy of a System 7.7 Conservative and Nonconservative Forces 7.8 Relationship Between Conserva- tive Forces and Potential Energy 7.9 Energy Diagrams and the Equilibrium of a System
ANSWERS TO QUESTIONS
Q7.1 (a) Positive work is done by the chicken on the dirt.
(b) The person does no work on anything in the environ- ment. Perhaps some extra chemical energy goes through being energy transmitted electrically and is converted into internal energy in his brain; but it would be very hard to quantify “extra.”
(c) Positive work is done on the bucket.
(d) Negative work is done on the bucket.
(e) Negative work is done on the person’s torso.
Q7.2 Force of tension on a ball moving in a circle on the end of a string. Normal force and gravitational force on an object at rest or moving across a level floor.
Q7.3 (a) Tension (b) Air resistance (c) The gravitational force does positive work in increasing speed on the downswing. It does negative work in decreasing speed on the upswing.
*Q7.4 Each dot product has magnitude 1⋅ 1 ⋅cos θ where θ is the angle between the two factors. Thus for (a) and (f ) we have cos 0 = 1. For (b) and (g), cos 45° = 0.707. For (c) and (h), cos 180° = −1. For (d) and (e), cos 90° = 0. The assembled answer is a = f > b = g > d = e > c = h.
Q7.5 The scalar product of two vectors is positive if the angle between them is between 0 and 90°. The scalar product is negative when 90 ° < θ < 180 °.
*Q7.6 (i) The force of block on spring is equal in magnitude and opposite to the force of spring on block. The answers are (c) and (e). (ii) The spring tension exerts equal-magnitude forces toward the center of the spring on objects at both ends. The answers are (c) and (e).
Q7.7 k ′ = 2 k. To stretch the smaller piece one meter, each coil would have to stretch twice as much as one coil in the original long spring, since there would be half as many coils. Assuming that the spring is ideal, twice the stretch requires twice the force.
Q7.8 No. Kinetic energy is always positive. Mass and squared speed are both positive. A moving object can always do positive work in striking another object and causing it to move along the same direction of motion.
153
Q7.9 Work is only done in accelerating the ball from rest. The work is done over the effective length of the pitcher’s arm—the distance his hand moves through windup and until release. He extends this distance by taking a step forward.
*Q7.10 answer (e). Kinetic energy is proportional to mass.
*Q7.11 answer (a). Kinetic energy is proportional to squared speed. Doubling the speed makes an object’s kinetic energy four times larger.
*Q7.12 It is sometimes true. If the object is a particle initially at rest, the net work done on the object is equal to its fi nal kinetic energy. If the object is not a particle, the work could go into (or come out of) some other form of energy. If the object is initially moving, its initial kinetic energy must be added to the total work to find the fi nal kinetic energy.
*Q7.13 Yes. The floor of a rising elevator does work on a passenger. A normal force exerted by a station- ary solid surface does no work.
*Q7.14 answer (c). If the total work on an object is zero in some process, its kinetic energy and so its speed must be the same at the final point as it was at the initial point.
*Q7.15 The cart’s fixed kinetic energy means that it can do a fixed amount of work in stopping, namely (6 N)(6 cm) = 0.36 J. The forward force it exerts and the distance it moves in stopping must have this fi xed product. answers: (i) c (ii) a (iii) d
Q7.16 As you ride an express subway train, a backpack at your feet has no kinetic energy as measured by you since, according to you, the backpack is not moving. In the frame of reference of someone on the side of the tracks as the train rolls by, the backpack is moving and has mass, and thus has kinetic energy.
*Q7.17 answer (e). (^) 4 00 1 2
. J = k ( 0 100. m)^2
Therefore k = 800 N m and to stretch the spring to 0.200 m requires extra work
∆ W = ( )( ) − =
800 0 200. 2 4 00. J 12 0. J
Q7.18 (a) Not necessarily. It does if it makes the object’s speed change, but not if it only makes the direction of the velocity change.
(b) Yes, according to Newton’s second law.
*Q7.19 (i) The gravitational acceleration is quite precisely constant at locations separated by much less than the radius of the planet. Answer: a = b = c = d
(ii) The mass but not the elevation affects the gravitational force. Answer: c = d > a = b
(iii) Now think about the product of mass times height. Answer: c > b = d > a
Q7.20 There is no violation. Choose the book as the system. You did work and the Earth did work on the book. The average force you exerted just counterbalanced the weight of the book. The total work on the book is zero, and is equal to its overall change in kinetic energy.
Q7.21 In stirring cake batter and in weightlifting, your body returns to the same conformation after each stroke. During each stroke chemical energy is irreversibly converted into output work (and inter- nal energy). This observation proves that muscular forces are nonconservative.
154 Chapter 7
156 Chapter 7
*P7.4 (^) Yes. Object 1 exerts some forward force on object 2 as they move through the same displacement.
By Newton’s third law, object 2 exerts an equal-size force in the opposite direction on object 1. In W = F ∆ r cos θ, the factors F and ∆ r are the same, and θ differs by 180°, so object 2 does −15.0 J of work on object 1. The energy transfer is 15 J from object 1 to object 2, which can be counted as a change in energy of −15 J for object 1 and a change in energy of +15 J for object 2.
Section 7.3 The Scalar Product of Two Vectors
P7.
A B ⋅ = (^) ( A (^) x i^ ˆ^ + A (^) y j ˆ^ + Az k ˆ^ (^) )⋅ (^) ( B (^) x i ˆ^ + B (^) y ˆ j^ + Bz k ˆ) A B ⋅ = (^) ( i i ⋅ (^) ) + (^) ( i ⋅ j (^) ) + (^) ( i k ⋅ )
A B A B A B
A
x x^ ˆ ˆ^ x y ˆ ˆ^ x z ˆ^ ˆ yy x y y y z z x
B A B A B
A B
j i j j j k
k
( ⋅) +^ ( ⋅) +^ ( ⋅ )
- ⋅ ˆˆ i^ k j ˆ^ ˆ^ k k ˆ^ ˆ A B
( ) +^ ( ⋅) +^ ( ⋅ ) ⋅ = +
A B A B
A B A B
z y z z x x y
yy + A Bz z
P7.6 A = 5 00. ; B = 9 00. ;^ θ =^ 50 0.° A B ⋅ = AB cos θ = ( 5 00. )( 9 00. )^ cos 50 0. °= 28 9.
P7.7 (a) W = ⋅ = F xx + F yy = ( )( ) ⋅ + −( )
F ∆ r 6 00. 3 00. N m 2 00. ( 1. 000 ) N m⋅ = 16 0. J
(b)^ θ =^
( ) + −
cos− cos−
..
1 1 2
F ∆ r
F ∆ r ( (( )) (( ) + ( ))
2 2 2 =^ °
P7.8 We must first fi nd the angle between the two vectors. It is:
θ = 360 ° − 118 ° − 90 0. ° − 132 ° =20 0.°
Then (^) F v ⋅ = Fv cos θ = ( 32 8. N )( 0 173. m s) cos 20 0.°
or
F v ⋅ =
N m s
J
s
P7.9 (a)
A = 3 00. ˆ i^ −2 00. ˆ j B = 4 00. i ˆ^ −4 00. ˆ j^ θ =^
( )( )
cos− cos−.^. =
..
1 1 12 0^ 8 00
A B
AB
(b)
B = 3 00. i ˆ^ − 4 00. ˆ j^ +2 00. k ˆ A = − 2 00. i ˆ^ +4 00. j ˆ^ cos
θ =
( )( )
A B
AB
θ = 156°
(c)
A = i ˆ^ − 2 00. ˆ j^ +2 00. k ˆ B = 3 00. j ˆ^ +4 00. k ˆ^ θ =^
= −^ +
cos −^ cos−.^. .
1 1 6 00^ 8 00
A B
AB ..
⎝⎜^
FIG. P7.
Energy of a System 157
P7.
A − B = (^) ( 3 00. i ˆ^ + ˆ j^ − k ˆ^ (^) ) − − +( i ˆ^ 2 00. j ˆ^ +5 00. k ˆ)
A B i j k C A B j
. ˆ^ ˆ^. ˆ
(.^ ˆ^. 0^0 k ˆ^^ )⋅^ ( 4 00.^ ˆ i^^ −^ ˆ j^^ −6 00.^ k ˆ^ ) =^0 + −(^ 2 00.^ ) + +(^ 18 0.) ==^ 16 0.
*P7.11 Let θ represent the angle between
A and
B. Turning by 25° makes the dot product larger, so the angle between
C and^
B must be smaller. We call it θ − 25°. Then we have
A 5 cos θ = 30 and A 5 cos (θ − 25°) = 35
Then A cos θ = 6 and A (cos θ cos 25° + sin θ sin 25°) = 7
Dividing, cos 25° + tan θ sin 25° = 7 6 tan θ = (7 6 − cos 25°)sin 25° = 0.
θ = 31.6°. Then the direction angle of A is 60° − 31.6° = 28.4°
Substituting back, A cos 31.6° = 6 so
A = 7.05 m at 28.4°
Section 7.4 Work Done by a Varying Force
P7.12 Fx = ( 8 x − 16 ) N
(a) See figure to the right
(b) W net = − (^ 2 00^ m^ )(16 0^ N^ ) + (^ m^ )(^ N) = − 2
.... 1 12 0. J
P7.13 (^) W Fdx i
f = (^) ∫ =area under curve from^ x^ i to^ x^ f
(a) (^) x (^) i = 0 x (^) f = 8 00. m
W = area of triangle ABC = ⎛⎝^1 ⎞⎠ AC × 2
altitude,
W 0 8 1 2 → = ⎛⎝ ⎞⎠ ×^ 8 00.^ m^ ×^ 6 00.^ N^ = 24 0. J
(b) x (^) i = 8 00. m x (^) f = 10 0. m
W = area of ∆ CDE = ⎛⎝ ⎞⎠ CE ×
altitude,
W 8 10
→ = ⎛⎝ ⎞⎠ × (2 00^.^ m^ ) × −(^ 3 00.^ N^ ) =^ −3 00. J
(c) W 0 (^) → 10 = W (^) 0 → 8 + W 8 → 10 = 24 0. + −( 3 00. ) = 21 0. J
P7.14 W^ d^ x^ y^ dx i
f = (^) ∫ ⋅ = (^) ∫( + ) ⋅
F r 4 i 3 j i 0
5 ˆ ˆ (^) N ˆ
m
0
5 2
0
5 N m N m J
m m ∫ (^ )^ xdx^ +^ = (^ )^ =
x (^).
FIG. P7.
0 0
0
0
FIG. P7.
Energy of a System 159
*P7.20 The spring exerts on each block an outward force of magnitude
Fs = kx = ( 3 85. N m (^) )( 0 08. m ) =0 308. N
Take the + x direction to the right. For the light block on the left, the vertical forces are given by Fg = mg = ( 0 25. kg)( 9 8. m s (^2) ) =2 45. N, (^) ∑ Fy = 0 , n − 2 45. N = 0 , n = 2 45. N. Similarly for the heavier block n = Fg = ( 0 5. kg)( 9 8. m s (^2) ) =4 9. N
(a) For the block on the left, (^) ∑ Fx = max , − 0 308. N = ( 0 25. kg ) a , a = −1 23. m s 2. For the heavier block, +0 308. N = ( 0 5. kg ) a , a = 0 616. m s 2.
(b) For the block on the left, f (^) k = μ k n = 0 1 2 45. (. N ) =0 245. N
F ma a a
∑ x = x − + = ( ) = −
m s 2 N kg 2 2 m s 2 if the force of static friction is noot too large.
For the block on the right, f (^) k = μ k n =0 490. N. The maximum force of static friction would be larger, so no motion would begin and the acceleration is zero.
(c) Left block: f (^) k = 0 462 2 45. (. N ) =1 13. N. The maximum static friction force would be larger, so the spring force would produce no motion of this block or of the right-hand block, which could feel even more friction force. For both a = 0.
*P7.21 Compare an initial picture of the rolling car with a final picture with both springs compressed K (^) i + (^) ∑ W = Kf. Work by both springs changes the car’s kinetic energy
K k x x k x x K
m
i i f i f f
i
- (^1) ( − ) + (^) ( − ) = 2
1 1
2 1
2 2 2
2 2
2
v^2 ++ − (^) ( (^) )( (^) )
2 N m m
N m
2
2
m
kg J 68.0 J 0
( ) =
( ) −^ −^ =
v
v
i
i
J
kg m s ( ) =.
P7.22 (a) W^ d
W x x
i
f = ⋅
= (^) ( + −
∫
F r
15 000 N 10 000 N m 25 000 2 N m (^2) )) °
∫ dx
W x x x
cos
. 0
0
0 600
2 3
m
0
0 600.
m
W = kJ + kJ − kJ = kJ
FIG. P7.
2.45 N
n Fs
FIG. P7.
F ( N)
continued on next page
160 Chapter 7
(b) Similarly,
W = ( )( ) + (^ )( ) 15 0 1 00 −
kN m kN m m ..^.
kN m m^3
kJ , larger by 29
( 2 )(^ )
W = ..6%
P7.23 The same force makes both light springs stretch.
(a) The hanging mass moves down by
x x x mg k
mg k
mg k k
1 2 1 2 1 2
1 5. kg 9.88 m s m N
m 1 800 N
(^2) m
⎝⎜^
=. × −
(b) We define the effective spring constant as
k
F
x
mg mg k k k k
( + )
− 1 1
1 2 1 2
1
m 1 200 NN
m 1 800 N
⎝⎜^
⎠⎟^
1
P7.24 See the solution to problem 7.23.
(a) x^ mg^ k k
1 2
Both springs stretch, so the load moves down by a larger amount than
it would if either spring were missing.
(b) k k k
− 1 1 1 2
1 The spring constant of the series combination is less than the smaller of
the two individual spring constants, to describe a less stiff system, that stretches by a larger extension for any particular load.
P7.25 (a) The radius to the object makes angle θ with the horizontal, so its weight makes angle θ with the negative side of the x -axis, when we take the x -axis in the direction of motion tangent to the cylinder. F ma F mg F mg
∑ x = x − = =
cos cos
θ θ
(b) W d i
f = (^) ∫ ⋅
F r
We use radian measure to express the next bit of displacement as dr = Rd θ in terms of the next bit of angle moved through:
W mg Rd mgR
W mgR mgR
= ( − ) =
∫ cos^ θ^ θ^ sinθ
π π 0
2 0
2
P7.26 (^) k F x [ ] = ⎡ ⎣⎢^
N
m
kg m s m
kg s
2 2
FIG. P7.
m g
162 Chapter 7
P7.32 (a) ∆ K = K (^) f − K (^) i = m (^) f − = (^) ∑ W =
v^2 0 (area under curve from x = 0 to x = 5 00. m)
v (^) f m
= 2 (^ ) = 2 7 50(^ ) = 4 00
area J 1 94 kg
. m s .
(b) ∆ K^ =^ K^ f −^ K^ i =^ m^ f −^ =^ ∑ W =
v^2 (^0) (area under curve from x = 0 to x = 10 0. m)
v (^) f m
( )
( )
area J kg m s
(c) ∆ K = K (^) f − K (^) i = m (^) f − = (^) ∑ W =
v^2 0 (area under curve from x = 0 to x = 15 0. m)
v (^) f m
( )
( )
area J kg m s
P7.33 Consider the work done on the pile driver from the time it starts from rest until it comes to rest at the end of the fall. Let d = 5.00 m represent the distance over which the driver falls freely, and h = 0 12. m the distance it moves the piling.
∑^ W^ = ∆ K^ :^ W^ gravity +^ W^ beam=^1 m^ f − mi 2
v^2 v^2
so ( mg ) ( h + d ) cos 0 ° + ( F (^) )( d ) cos 180 °= 0 − 0 Thus,
F mg h d d
= (^ )^ ( + ) = ( 2 100 )( 9 80 )( 5 12 ) 0 1
kg. m s 2. m
. 2 20
m
=. × N
The force on the pile driver is upward.
*P7.34 (a) We evaluate the kinetic energy of the cart and the work the cart would have to do to plow all the way through the pile. If the kinetic energy is larger, the cart gets through. K = (12) mv^2 = (12)(0.3 kg)(0.6 ms)^2 = 0.054 J The work done on the cart in traveling the whole distance is the net area under the graph, W = (2 N) (0.01 m) + [(0 − 3 N)2](0.04 m) = 0.02 J − 0.06 J = −0.04 J
The work the cart must do is less than the original kinetic energy, so the cart does get through all the sand. (b) The work the cart does is +0.04 J, so its final kinetic energy is the remaining 0.054 J − 0.04 J = 0.014 J. Another way to say it: from the work-kinetic energy theorem, Ki + W = Kf 0.054 J – 0.04 J = 0.014 J = (12)(0.5 kg) vf^2 vf = [2(0.014 kg⋅m^2 s 2 )(0.3 kg)] 1 ^2 = 0.306 m s
*P7.35 (a) K (^) i + (^) ∑ W = K (^) f = m f
v^2
0 1 2
- (^) ∑ W = (^) (15 0. × 10 −^3 kg)( 780 m s) 2 = 4 56. kJ
(b) F
W
r
×
( ) °
∆ cos
cos
θ
3 J
0.720 m kN
continued on next page
Energy of a System 163
(c) a x
f i f
= (^ ) − ( )
v^2 v^2 2
m s m
km s 2 .
(d) (^) ∑ F = ma = (^) ( 15 × 10 − 3 kg) (^) ( 422 × 10 3 m s (^2) ) = 6 34. kN
(e) The forces are the same. The two theories agree.
P7.36 (a) v^ f =^ 0 096 3.^ ( ×^10 8 m s) =^ 2 88.^ ×^107 m s
K (^) f = 1 m f = (^) ( × − ) ( × ) = 2
v^2^ 9 11. 10 31 kg 2 88. 10 7 m s 2 3 ..78 × 10 −^16 J
(b) K (^) i + W = Kf : 0 + F ∆ r cos θ= K (^) f F (0 028. m ) cos 0 ° = 3 78. × 10 −^16 J F = 1 35. × 10 − 14 N
(c) (^) ∑ F = ma ; a
F
m
= = ×
×
∑ (^) = × − −
14 31
N.
kg
mm s 2
(d) v (^) xf = vxi + a tx 2 88. × 10 7 m s = 0 + (^) (1 48. × 1016 m s (^2) ) t t = 1 94. × 10 − 9 s
Check: x f = x i + ( xi + xf ) t
v v
0 028 0
. m = + ( 0 + 2 88. × 10 7 m s) t
t = 1 94. × 10 − 9 s
Section 7.6 Potential Energy of a System
P7.37 (a) With our choice for the zero level for potential energy when the car is at point B, U B = 0
When the car is at point A, the potential energy of the car-Earth system is given by U (^) A = mgy
where y is the vertical height above zero level. With 135 ft = 41.1 m, this height is found as: y = ( 41 1. m ) sin 40 0. ° = 26 4. m Thus, U A= ( 1 000 kg (^) )( 9 80. m s^2 )( 26 4. m ) = 2 59. × 10 5 J
The change in potential energy as the car moves from A to B is U (^) B − U A = 0 − 2 59. × 10 5 J = − 2 59. × 105 J
FIG. P7.
continued on next page
Energy of a System 165
P7.40 (a) W = (^) ∫ ⋅ d
F r and if the force is constant, this can be written as
W = ⋅ ∫ d = ⋅ ( f − i )
F r F r r , which depends only oon end points, not path.
(b) W = (^) ∫ ⋅ d = (^) ∫( + )⋅ (^) ( dx + dy (^) ) = ( ) dx
F r 3 i ˆ^4 ˆ j^ i ˆ^ j ˆ^ 3 00. N 00
5 00
0
5 00 4 00
.. .
m m ∫ + (^ N)^ ∫ dy
W = ( 3 00. N ) x 5 00 0.^ m^ + ( 4 00. N) y 5 00 0. m= 15 0. J+ 2 20 0. J = 35 0. J
The same calculation applies for all paths.
P7.41 (a) The work done on the particle in its first section of motion is
WOA = (^) ∫ dx ˆ^ ⋅ (^) ( y ˆ^ + x ˆ) = ∫ ydx
.. i 2 i^2 j 2 0
5 00
0
m 5 00m
and since along this path, (^) y = 0 WOA = 0
In the next part of its path W (^) AC = (^) ∫ dy ˆ^ ⋅ (^) ( y ˆ^ + x ˆ) = ∫ x dy
.. j 2 i^2 j 0
5 00 2 0
m 5 00m
For x = 5 00. m, W (^) AC = 125 J
and WOAC = 0 + 125 = 125 J
(b) Following the same steps, WOB^ =^ ∫ dy ˆ^^ ⋅^ ( y^ ˆ^ + x^ ˆ) = ∫ x dy
.. j 2 i^2 j 0
5 00 2 0
m 5 00m
since along this path, x = 0 , WOB = 0
W (^) BC = (^) ∫ dx ˆ^ ⋅ (^) ( y ˆ^ + x ˆ) = ∫ ydx
.. i 2 i^2 j 2 0
5 00
0
m 5 00m
since y = 5 00. m, (^) W (^) BC = 50 0. J
WOBC = 0 + 50 0. = 50 0. J
(c) WOC = (^) ∫ ( dx ˆ i^ + dy ˆ j^ )⋅ (^) ( 2 y ˆ i^ + x^2 ˆ j ) = (^) ∫( 2 ydx + x dy^2 )
Since x = y along OC , WOC = (^) ∫ ( 2 x + x^2 ) dx = 66 7 0
5 00. .
m J
(d) F is nonconservative since the work done is path dependent.
P7.42 Along each step of motion, the frictional force is opposite in direction to the incremental displacement, so in the work cos 180° = −1.
(a) W = (3 N)(5 m)(−1) + (3 N)(5 m)(−1) = −30.0 J
(b) The distance CO is (5 2 + 52 )^1 ^2 m = 7.07 m
W = (3 N)(5 m)(−1) + (3 N)(5 m)(−1) + (3 N)(7.07 m)(−1) = −51.2 J
(c) W = (3 N)(7.07 m)(−1) + (3 N)(7.07 m)(−1) = −42.4 J
(d) The force of friction is a nonconservative force.
166 Chapter 7
Section 7.8 Relationship Between Conservative Forces and Potential Energy
P7.43 (a) W F dx x dx x = (^) x = ( + ) = + x
⎝⎜^
∫ ∫ (^) ⎠⎟
1
5 00 (^2)
1
. m 5 00. m = 25 0. + 20 0. − 1 00. − 4 00. = 40 0. J
(b) ∆ K + ∆ U = 0 ∆ U = − ∆ K = − W = −40 0. J
(c) ∆ K = K (^) f − mv^1
2 2
K K
m f =^ ∆^ +^ =
v 12 2
62 5. J
P7.44 (a) (^) U Ax Bx dx Ax^ Bx
x = − (^) ∫ (− +^2 ) = − 0
2 3 2 3
(b) ∆ U Fdx
A
⎡⎣( ) − ( )⎤⎦ ∫ − 2 00
3 00^2 2 00^2
.^2
m
3.00 m (^) BB A B
⎡⎣( ) − ( )⎤⎦ =. (^) −.
∆ K = ⎛⎝− A + B ⎞⎠
P7.45 U r
A
r
F
U
r
d dr
A
r
A
r (^) r
= − ⎛⎝ ⎞⎠ = (^2). If A is positive, the positive value of radial force
indicates a force of repulsion.
P7.46 F
U
x
x y x x x = − x y^ x y
∂ (^) ( − ) ∂
= − (^) ( − ) = −
3 2 2
F U
y
x y x y y = − ∂ x^ x ∂
∂ ( − ) ∂
= − (^) ( − ) = −
3 3 3
Thus, the force acting at the point ( x , y ) is
F = Fx i^ ˆ^ + Fy ˆ j^ = (^) ( 7 − 9 x y^2 (^) ) i ˆ^ − 3 x^3 ˆ^ j.
Section 7.9 Energy Diagrams and the Equilibrium of a System
P7.47 (a) Fx is zero at points A, C and E; Fx is positive at point B and negative at point D.
(b) A and E are unstable, and C is stable.
(c)
FIG. P7.
A
B
C
D
E (^) x (m)
Fx
168 Chapter 7
P7.51 At start, v = ( 40 0. m s ) cos 30 0. ° i ˆ^ + ( 40 0. m s)sin 30 0. ° j ˆ
At apex, v = ( 40 0. m s) cos 30 0. ° i ˆ^ + 0 j ˆ^ = ( 34 6. m s) i ˆ And K = m = ( )( ) =
v^2^ 0 150. kg 34 6. m s 2 90 0. J
P7.52 (a) We write F ax a a
b b b
= ( ) = ( )
=
N m N m
ln ln. ln ln.
b b
b
b 44
1 80 4
( )
= ×
b
a N 0.129 m
N m 1..8^ = a
(b) (^) W = Fdx = × x dx
∫ ∫ 0
0 25 4 1 8 0
0 25 4 01 10
. . . .
m
m (^) N m
4 4 01 10 2 8
2 8
0
0 25
. 4 .
.. × = ×
N
m
N
1.8 m
m
x 88
m
J
( ).
*P7.53 (a) We assume the spring is in the horizontal plane of the motion. The radius of the puck’s motion is 0.155 m + x
The spring force causes the puck’s centripetal acceleration: (4.3 Nm) x = F = mv^2 r = m (2π r T)^2 r = 4 π 2 m r (1.3 s)^2
(4.3 kgs^2 ) x = (23.4s^2 ) m (0.155 m + x )
4.3 kg x = 3.62 m m + 23.4 m x
4.3000 kg x − 23.360 m x = 3.6208 m m
x = (3.62 m )(4.3 kg − 23.4 m ) meters
(b) x = (3.62 m 0.07 kg)(4.30 kg − 23.4 [0.07 kg]) = 0.0951 m a nice reasonable extension
(c) We double the puck mass and find x = (3.6208 m 0.14 kg)/(4.30 kg − 23.360 [0.14 kg]) = 0.492 m more than twice as big!
(d) x = (3.62 m 0.18 kg)(4.30 kg − 23.4 [0.18 kg]) = 6.85 m We have to get a bigger table!!
(e) When the denominator of the fraction goes to zero, the extension becomes infinite. This hap- pens for 4.3 kg − 23.4 m = 0; that is for m = 0.184 kg. For any larger mass, the spring cannot constrain the motion. The situation is impossible.
(f) The extension is directly proportional to m when m is only a few grams. Then it grows faster and faster, diverging to infinity for m = 0.184 kg.
Energy of a System 169
*P7.54 (a) A time interval. If the interaction occupied no time, the force exerted by each ball on the other would be infinite, and that cannot happen.
(b) k = | F || x | = 16 000 N0.000 2 m = 80 MNm (c) We assume that steel has the density of its main constituent, iron, shown in Table 14.1. Then its mass is ρ V = ρ (43) π r^3 = (4π3)(7860 kgm 3 )(0.0254 m2) 3 = 0.0674 kg and K = (12) mv^2 = (12) (0.0674 kg)(5 ms) 2 = 0.843 J ≈ 0.8 J (d) Imagine one ball running into an infinitely hard wall and bouncing off elastically. The origi- nal kinetic energy becomes elastic potential energy 0.843 J = (12) (8 × 107 Nm) x^2 x = 0.145 mm ≈ (^) 0.15 mm
(e) The ball does not really stop with constant acceleration, but imagine it moving 0.145 mm forward with average speed (5 ms + 0) 2 = 2.5 ms. The time interval over which it stops is then 0.145 mm(2.5 ms) = 6 × 10 −^5 s ∼ 10 −^4 s
*P7.55 The potential energy at point x is given by 5 plus the negative of the work the force does as a particle feeling the force is carried from x = 0 to location x. dU Fdx dU e xdx U
U x = − (^) ∫∫ 5 = − (^) ∫ 08 −^2 − 5 = −( ⎡⎣− ⎤⎦) (^) ∫ (− )
= − (^) ( ⎡⎣− ⎤⎦)
−
−
2 0
2
e dx
U e
x x
x 00
x (^) = 5 + 4 e − 2 x (^) − 4 1⋅ = 1 + 4 e − 2 x
The force must be conservative because the work the force does on the object on which it acts depends only on the original and final positions of the object, not on the path between them.
P7.56 (a)
F = − d ( − + + ) i = (^) ( − − ) i dx
x^3 2 x^2 3 x ˆ^3 x^24 x 3 ˆ
(b) F = 0
when x = 1 87. and−0 535.
(c) The stable point is at
x = −0 535. point of minimum U ( x )
The unstable point is at
x = 1 87. maximum in U ( x )
FIG. P7.
Energy of a System 171
P7.59 We evaluate by 375 12 8^3 3 75
23 7 (^) dx ∫. (^) x +. x
. calculating
( ) ( ) + ( )
(( )
and 375 0 100 12 9 3 75 12 9
( ) ( ) + ( )
(( )
The answer must be between these two values. We may find it more precisely by using a value for ∆ x smaller than 0.100. Thus, we find the integral to be 0 799. N m⋅.
P7.60 (a) F N L mm F N Lmm
( ) ( ) ( ) ( ) 1 15.
To draw the straight line we use all the points listed and also the origin. If the coils of the spring touched each other, a bend or nonlinearity could show up at the bottom end of the graph. If the spring were stretched “too far,” a nonlinearity could show up at the top end. But there is no visible evidence for a bend in the graph near either end.
(b) By least-square fitting, its slope is
0 125. N mm ± 2 % = 125 N m ± 2 %
In F = kx , the spring constant is k F x
= , the same as the slope of the F -versus- x graph.
(c) F = kx = ( 125 N m)( 0 105. m ) = 13 1. N
ANSWERS TO EVEN PROBLEMS
P7.2 (a) 3 28. × 10 −^2 J (b) − 3 28. × 10 − 2 J
P7.4 Yes. It exerts a force of equal magnitude in the opposite direction that acts over the same distance. −15.0 J
P7.6 28.
P7.8 5.33 Js
FIG. P7.
172 Chapter 7
P7.10 16.
P7.12 (a) see the solution (b) −12 0. J
P7.14 50.0 J
P7.16 (a) 575 N m (b) 46.0 J
P7. 18 (a) 1.13 kNm (b) 51.8 cm
P7.20 (a) −1.23 ms^2 and +0.616 ms 2 (b) −0.252 ms 2 and 0 (c) 0 and 0
P7.22 (a) 9.00 kJ (b) 11.7 kJ, larger by 29.6%
P7.24 (a) mg k
mg 1 k 2
1 k k
−
P7.26 kgs^2
P7. 28 If the weight of the first tray stretches all four springs by a distance equal to the thickness of the tray, then the proportionality expressed by Hooke’s law guarantees that each additional tray will have the same effect, so that the top surface of the top tray will always have the same elevation. 316 Nm. We do not need to know the length and width of the tray.
P7.30 (a) 33.8 J (b) 135 J
P7.32 (a) 1 94. m s (b) 3 35. m s (c) 3 87. m s
P7. 34 (a) yes. Its kinetic energy as it enters the sand is sufficient to do all of the work it must do in plowing through the pile. (b) 0.306 ms
P7.36 (a) 3 78. × 10 −^16 J (b) 1 35. × 10 −^14 N (c) 1 48. × 10 +^16 m s 2 (d) 1.94 ns
P7.38 (a) 800 J (b) 107 J (c) 0
P7.40 (a) see the solution (b) 35.0 J
P7. 42 (a) −30.0 J (b) −51.2 J (c) −42.4 J (d) The force of friction is a nonconservative force.
P7.44 (a) Ax^2 2 − Bx^3 3 (b) ∆ U = 2.5 A −6.33 B ; ∆ K = −2.5 A + 6.33 B
P7.46 (^) ( 7 − 9 x y^2 ) (^) ˆ i^ − 3 x^3 ˆ j
P7.48 see the solution
P7.50 k 1 xmax^2 2 + k 2 xmax^3 3
P7.52 (a) a m
1 8
.
kN ; b = 1 80. (b) 294 J
P7. 54 (a) A time interval. If the interaction occupied no time, the force exerted by each ball on the other would be infi nite, and that cannot happen. (b) 80 MNm (c) 0.8 J. We assume that steel has the same density as iron. (d) 0.15 mm (e) 10−^4 s
