Download Solutions to Problems: Work and Energy

Solutions to Problems: Work and Energy
P162: 37, 49, 58, 67, 76. 37. The force needed to hold a particular spring compressed an amount x from its
normal length is given by F = kx + ax3 + bx 4 . How much work must be done to
compressed an amount X, starting from x = 0?
Solution: According to the definition, we get
X
W = ∫ Fdx = ∫
0
X
0
( kx + ax
3
+ bx 4 )dx =
1 2 1
1
kX + aX 4 + bX 5 .
2
4
5
49. A force of 6.0 N is used to accelerate a mass of 1.0 kg from rest for a distance of
12 m. The force is applied along the direction of travel. The coefficient of kinetic
friction is 0.30. What is the work done (a) by the applied force? (b) by friction? (c)
What is the kinetic energy at the 12-m mark?
Solution: The mass undergoes one-dimensional motion, and thus, (a) the work done
by the applied force is W1 = FS = 6.0 × 12 J = 72 J . And (b) the work done by friction
can be obtained as W2 = − f k S = − μ k mgS = −0.3 × 1.0 × 9.8 × 12 J = −35 J . (c) From
work- energy principle, the final kinetic energy is K 2 = K1 + Wnet = ( 72 − 35 ) J = 37 J .
58. We usually neglect the mass of a spring if it is a small compared to the mass
attached to the spring. But in some applications, the mass of the spring must be taken
into account. Consider a spring of unstretched length L and mass MS. This mass is
uniformly distributed along the length of the spring. A mass m is attached to the end
of the spring. One end of the spring is fixed and the mass is allowed to vibrate
horizontally without friction (see Fig. 7-30). Each point on the spring moves with a
velocity proportional to the distance from that point to the fixed end. For example, if
the mass on the end moves with speed v, the midpoint of the spring moves with speed
v/2. Show the kinetic energy of the mass plus spring when the mass is moving with
velocity v is K = Mv2/2, where M = m + MS/3 is the “effective mass” of the system.
Solution: Assume the length of the spring is X in this case the mass is moving with
speed v. Choose a segment element dx in the spring, and it defines a mass element
dm = ρ dx =
MS
1
x
dx , and the kinetic energy reads dK x = ( dm ) vx2 , with vx = v . Thus
2
X
X
K
the kinetic energy of the spring is K S = ∫ dK x =
0
M sv2
2X 3
∫
X
0
x 2 dx =
1
M s v 2 , and the total
6
kinetic energy of the system is
Kt = K S + K m =
1
1
1
1
M S v 2 + mv 2 = Mv 2 , M = M S + m, QED.
6
2
2
3
(
)
67. A force F = 10.0i + 9.0 j + 12.0k kN acts on a small object of mass 100 g. If the
displacement of the object is d = ( 5.0i + 4.0 j ) m, find the work done by the force.
What is the angle between F and d .
Solution: The work done by the force is W = F id = ( 50 + 36 ) kJ = 86 kJ . And the
angle reads θ = arccos
F id
86
= arccos
= arccos 0.746, or θ = 42° .
Fd
325 × 41
76. A simple pendulum consists of a small object of mass m (the bob) suspended by a
cord of length L (Fig. 7-33) of negligible mass. A force is applied in the horizontal
direction, moving the bob very slowly so the acceleration is essentially zero. (Note
that the magnitude of the force needed to vary with the angle that the cord makes with
the vertical at any moment.) (a) Determine the work done by this force, to move the
pendulum from θ = 0 to θ = θ 0 . (b) Determine the work done by the gravitational force
on the bob, FG = mg , and the work done by the force that the cord exerts on the bob.
Solution: (a) According to conservation law of the energy, the work done by this force
should equal to the change of the potential of the bob:
W = mg Δh = mgL (1 − cos θ 0 ) .
(b) The kinetic energy of the bob is unchanged during the course and the work done
by the tension is zero since the displacement is always perpendicular to the
displacement. And the work done by the gravitational force is
W = −mg Δh = − mgL (1 − cos θ 0 ) .
P192: 6, 7, 32, 67, 70, 79. 6. If U = 3x 2 + 2 xy + 4 y 2 z , what is the force?
Solution: The force is the negative gradient of the potential energy function, So
⎛ ∂U ∂U ∂U ⎞
2
F = −∇U = − ⎜
,
,
⎟ = − ( 6 x − 2 y, 2 x + 8 yz, 4 y ) ( N ) . .
x
y
z
∂
∂
∂
⎝
⎠
7. A particular spring obeys the force law F = ( − kx + ax3 + bx 4 ) i . (a) Is this force
conservative? Explain why or why not. (b) If it is conservative, determine the form of
the potential energy function.
Solution: (a) The force is conservative, since the work done by it is independent of
x2
1
1
1
the path: W = Wx = ∫ Fx dx = − kx2 + ax4 + bx5
x1
2
4
5
x2
x1
. (b) The potential energy function
1
1
1
is of the form: U2 −U1 = −Wx = − kx2 + ax4 + bx5
2
4
5
x1
x2
1
1
1
⇒ U ( x) = kx2 − ax4 − bx5 ,
2
4
5
where we take U ( x = 0 ) = 0 .
32. Consider the track shown in Fig. 8-32. The section AB is one quadrant of a circle
of radius 2.0 m and is frictionless. B to C is horizontal span 3.0m long with a
coefficient of kinetic friction μk = 0.25 . The section CD under the spring is
frictionless. A block of mass 1.0 kg is released from rest at A. After sliding on the tract
it is observed to compress the spring by 0.20 m. Determine: (a) the velocity of the
block at point B; (b) the work done by friction as the block slides from B to C; (c) the
velocity of the block at point C; (d) the stiffness constant k for the spring.
Solution: (a) From the work-energy principle, the change of kinetic energy of the
block is K f − K i = K f = mgh =
1 2
mv f ⇒ v f = 2 gh = 6.3 m/s .
2
(b) The force of kinetic friction is f k = μk mg , and thus the work done by friction reads
W = − f k L = − μk mgL = − 7.4 J .
(c) According to work-energy principle, we get
K f − Ki = W f ⇒ K f = Ki − f k L ⇒ v f =
( Ki − f k L )
2m = 4.9 m/s .
(d) According to the problem, we learn
1 2
kx = K f ⇒ k = 2 K f x 2 = 6.1×102 N/m .
2
67. The position of a 280-g object is given (in meters) by x = 5.0t 3 − 8.0t 2 − 30t , where
t is in seconds. Determine the net rate of work done on this object (a) at t = 2.0 s and
(b) t = 4.0 s. (c) What is the average net power input during the interval from t = 0 s to
t = 2.0 s, and in the interval from t = 2.0 s to t = 4.0 s?
Solution: The velocity and acceleration of the object are vx =
dx
= 15t 2 −16t − 30 ( m/s) ,
dt
d 2x
and a x = 2 = 30t − 16 ( m/s 2 ) , and the net rate of work done on this object
dt
reads P = ma iv = 63t 3 − 100.8t 2 − 90.16t + 67.2 ( J/s ) , so (a) P
P t =4.0s = 4.3×103 J/s . (c) Since vx
one can get kinetic energy as K1
t =0s
t =0s
= −30 m/s, vx
= 126 J, K2
t =2.0s
t =2.0s
t = 2.0s
= −25 J/s ; (b)
= −2 m/s, and vx
= 0.56 J, and K3
t =4.0s
t =4.0s
= 146 m/s ,
= 2.98×103 J .
According to work-energy principle, we have the work done by the net force during
the interval from t = 0 s to t = 2.0 s and in the interval from t = 2.0 s to t = 4.0 s read:
W1 = K2 − K1 = −125 J, W2 = K3 − K2 = 2.97 ×103 J , therefore, the average power read:
P1 = W1 ( t2 − t1 ) = −63 W, P2 = W2 ( t3 − t2 ) = 1.5 × 103 W .
70. The potential energy of the two atoms in a diatomic molecule can be written
as U ( r ) = − a / r 6 + b / r12 , where r is the distance between the two atoms and a and b
are positive constants. (a) At what values of r is U(r) a minimum? A maximum? (b) At
what values of r is U ( r ) = 0 ? (c) Plot U(r) as a function of r from r = 0 to r at a value
large enough for all the features in (a) and (b) to show. (d) Describe the motion of one
atom with respect to the second atom when E < 0. (e) Let F be the force one atom
exerts on the other. For what values of r is F > 0, F < 0, and F = 0? (f) Determine F
as a function of r.
Solution: (a) From ∂U ∂r r0 = 0, one can get r0 = ( 2b a ) , and
16
∂2U ∂2r
r0
= 6r0−14 ( 26b − 7ar06 ) = 72br0−14 > 0,
So at r0 = ( 2b a ) , U is a minimum. For there is only one stationary point, U
16
approaches its maximum as r approaches zero.
(b) From U ( r1 ) = − a / r16 + b / r112 = 0 , one can figure out r1 = ( b a )
16
.
(c) Omitted.
∂U
= 6a r 7 − 12b / r13 , ar = 6a m1r 7 − 12b / m1r13 ,
∂r
and E = K +U < 0, the atom must oscillate between the turning points where E = U.
∂U
(e) From Fr = −
= 6a r 7 − 12b / r13 , we can easily find that
∂r
(d) U ( r ) = − a / r 6 + b / r12 ⇒ Fr = −
Fr = 0, for r = r0 =
2b
; Fr > 0 for r < r0 , Fr < 0 for r > r0 .
a
(f) Fr = − ∂U ∂r = 6a r 7 − 12b / r13 .
79. A ball is attached to a horizontal cord of length L whose other end is fixed, Fig.
8-36. (a) If the ball is released, what will be its speed at the lowest point of its path? (b)
a ped is located a distance h directly below the point of attachment of the cord. If h =
0.80 L, what will be the speed of the ball when it reaches the top of its circular path
about the peg?
Solution: (a) With the conservation law of mechanical law in mind, the speed can be
obtained as
v = 2 K m = 2mgL m = 2 gL .
(b) Similar to case (a), one can get
v = 2 K m = 2mg ( L − 2 R ) m = 2 g ( 2h − L ) = 1.2 gL .