Download Chapter 7 Solutions

PH211 Chapter 7 – Solutions
7.4.
IDENTIFY: The energy from the food goes into the increased gravitational potential
energy of the hiker. We must convert food calories to joules.
SET UP: The change in gravitational potential energy is Ugrav  mg ( yf  yi ), while the
increase in kinetic energy is negligible. Set the food energy, expressed in joules, equal to
the mechanical energy developed.
EXECUTE: (a) The food energy equals mg ( yf  yi ), so
yf  yi 
(140 food calories)(4186 J/1 food calorie)
(65 kg)(980 m/s2 )
 920 m.
(b) The mechanical energy would be 20% of the results of part (a), so
y  (020)(920 m)  180 m.
EVALUATE: Since only 20% of the food calories go into mechanical energy, the hiker
needs much less of climb to turn off the calories in the bar.
7.5.
IDENTIFY and SET UP: Use energy methods. Points 1 and 2 are shown in Figure 7.5.
(a) K1  U1  Wother  K2  U2. Solve for K 2 and then use K2  12 mv22 to obtain v2 .
Wother  0
(The only force on the ball
while
it is in the air is gravity.)
K1  12 mv12 ; K2  12 mv22
U1  mgy1, y1  22.0 m
U 2  mgy2  0,
since y2  0
for our choice of coordinates.
Figure 7.5
EXECUTE:
1 mv 2
1
2
 mgy1  12 mv22
v2  v12  2 gy1  (120 m/s)2  2(980 m/s2 )(220 m)  240 m/s
EVALUATE: The projection angle of 531 doesn’t enter into the calculation. The kinetic
energy depends only on the magnitude of the velocity; it is independent of the direction
of the velocity.
(b) Nothing changes in the calculation. The expression derived in part (a) for v2 is
independent of the angle, so v2  240 m/s, the same as in part (a).
(c) The ball travels a shorter distance in part (b), so in that case air resistance will have
less effect.
7.16.
IDENTIFY: We treat the tendon like a spring and apply Hooke’s law to it. Knowing the
force stretching the tendon and how much it stretched, we can find its force constant.
SET UP: Use Fon tendon  kx. In part (a), Fon tendon equals mg, the weight of the object
suspended from it. In part(b), also apply Uel  12 kx2 to calculate the stored energy.
Fon tendon (0250 kg)(980 m/s 2 )

 199 N/m.
x
00123 m
138 N

 0.693m  69.3 cm; Uel  12 (199 N/m)(0.693 m)2  47.8 J.
199 N/m
EXECUTE: (a)
(b)
x
Fon tendon
k
k
EVALUATE: The 250 g object has a weight of 2.45 N. The 138 N force is much larger
than this and stretches the tendon a much greater distance.
IDENTIFY: The spring force is conservative but the force of friction is nonconservative. Energy is conserved
during the process. Initially all the energy is stored in the spring, but part of this goes to kinetic energy, part
remains as elastic potential energy, and the rest does work against friction.
7.26.
SET UP: Energy conservation: K1  U1  Wother  K2  U 2 , the elastic energy in the spring is U  12 kx 2 , and
the work done by friction is Wf   fk s  k mgs.
EXECUTE: The initial and final elastic potential energies are U1  12 kx12  12 (840 N/m)(00300 m)2  0378 J
and U 2  12 kx22  12 (840 N/m)(00100 m)2  00420 J. The initial and final kinetic energies are K1  0 and
K2  12 mv22 . The work done by friction is
Wother  W fk   f k s  k mgs  (040)(250 kg)(98 m/s2 )(00200 m)  0196 J. Energy conservation gives
K2  12 mv22  K1  U1  Wother  U 2  0378 J  (0196 J)  00420 J  0140 J. Solving for v2 gives
v2 
2K2
2(0140 J)

 0335 m/s.
m
250 kg
EVALUATE: Mechanical energy is not conserved due to friction.
IDENTIFY: Some of the initial gravitational potential energy is converted to kinetic energy, but some of it is
lost due to work by the nonconservative friction force.
SET UP: The energy of the box at the edge of the roof is given by: Emech, f  Emech, i  f k s. Setting yf  0 at
7.32.
this point, yi  (425 m) sin36  250 m Furthermore, by substituting Ki  0 and Kf  12 mvf 2 into the
conservation equation,
1 mv 2
f
2
 mgyi  f k s or vf  2 gyi  2 f k sg/w  2 g ( yi  f k s/w).
EXECUTE: vf  2(980 m/s2 ) (250 m)  (22 0 N)(4 25 m)/(85 0 N)  524 m/s.
EVALUATE: Friction does negative work and removes mechanical energy from the system. In the absence of
friction the final speed of the toolbox would be 700 m/s
7.35
.
IDENTIFY: Apply Eq. (7.16).
SET UP: The sign of Fx indicates its direction.
dU
 4 x3  (48J/m4 ) x3. Fx (0.800m)  (4.8J/m4 )(0.80m)3  2.46N. The force
dx
is in the  x-direction.
EVALUATE: Fx  0 when x  0 and Fx  0 when x  0, so the force is always directed towards the origin.
EXECUTE: Fx  
7.39.
IDENTIFY and SET UP: Use Eq. (7.17) to calculate the force from U. At equilibrium F  0.
(a) EXECUTE: The graphs are sketched in Figure 7.39.
U
a
12
r
F 

b
r6
dU
12a 6b
  13  7
dr
r
r
Figure 7.39
(b) At equilibrium F  0, so
dU
0
dr
F  0 implies
12a
13
r

6b
r7
0
6br 6  12a; solution is the equilibrium distance r0  (2a/b)1/ 6
U is a minimum at this r; the equilibrium is stable.
(c) At r  (2a/b)1/6 , U  a/r12  b/r 6  a(b/2a)2  b(b/2a)  b2/4a.
At r  , U  0 The energy that must be added is U  b2/4a.
(d) r0  (2a/b)1/6  113 1010 m gives that
2a/b  2082 1060 m6 and b/4a  2402 1059 m6
b2/4a  b(b/4a)  154 1018 J
b(2402 1059 m6 )  154 1018 J and b  6411078 J  m6.
Then 2a/b  2082 1060 m6 gives a  (b/2)(2082 1060 m6 ) 
1 (641 1078
2
J  m6 ) (2082 1060 m6 )  667 10138 J  m12
EVALUATE: As the graphs in part (a) show, F (r ) is the slope of U (r ) at each r. U (r ) has a minimum where
F  0.
7.41 .
IDENTIFY: Apply F  ma to the bag and to the box. Apply Eq. (7.7) to the motion of the system of the box
and bucket after the bag is removed.
SET UP: Let y  0 at the final height of the bucket, so y1  200 m and y2  0 . K1  0. The box and the
bucket move with the same speed v, so K2  12 (mbox  mbucket )v2. Wother   f k d , with d  200 m and
f k  k mbox g. Before the bag is removed, the maximum possible friction force the roof can exert on the box is
(0700)(800 kg  500 kg)(980 m/s 2 )  892 N. This is larger than the weight of the bucket (637 N), so before
the bag is removed the system is at rest.
EXECUTE: (a) The friction force on the bag of gravel is zero, since there is no other horizontal force on the bag
for friction to oppose. The static friction force on the box equals the weight of the bucket, 637 N.
(b) Eq. (7.7) gives mbucket gy1  f k d  12 mtot v 2 , with mtot  1450 kg. v 
v
2
(mbucket gy1  k mbox gd ).
mtot
2
[(650 kg)(980 m/s 2 )(200 m)  (0400)(800 kg)(980 m/s2 )(200 m)].
1450 kg
v  299 m/s.
EVALUATE: If we apply F  ma to the box and to the bucket we can calculate their common acceleration a.
Then a constant acceleration equation applied to either object gives v  299 m/s, in agreement with our result
obtained using energy methods.
7.43.
IDENTIFY: Use the work-energy theorem, Eq. (7.7). The target variable  k will be a factor in the work done
by friction.
SET UP: Let point 1 be where the block is released and let point 2 be where the block stops, as shown in Figure
7.43.
K1  U1  Wother  K2  U 2
Work is done on the block by
the spring and by friction,
so Wother  W f and U  U el .
Figure 7.43
EXECUTE: K1  K2  0
U1  U1,el  12 kx12  12 (100 N/m)(0200 m)2  200 J
U 2  U 2,el  0, since after the block leaves the spring has given up all its stored energy
Wother  W f  ( fk cos )s  k mg  cos  s  k mgs, since   180 (The friction force is directed opposite to
the displacement and does negative work.)
Putting all this into K1  U1  Wother  K2  U 2 gives
U1,el  W f  0
k mgs  U1,el
k 
U1,el
mgs

2.00 J
(050 kg)(980 m/s 2 )(100 m)
 041.
EVALUATE: U1,el  W f  0 says that the potential energy originally stored in the spring is taken out of the
system by the negative work done by friction.
7.45.
IDENTIFY: The mechanical energy of the roller coaster is conserved since there is no friction with the track.
We must also apply Newton’s second law for the circular motion.
SET UP: For part (a), apply conservation of energy to the motion from point A to point B:
K B  Ugrav,B  K A  Ugrav,A with K A  0. Defining yB  0 and y A  130 m, conservation of energy
becomes
1
mvB 2
2
 mgy A or vB  2 gy A . In part (b), the free-body diagram for the roller coaster car at point B
is shown in Figure 7.45. Fy  ma y gives mg  n  marad , where arad  v 2 /r. Solving for the normal force
 v2

gives n  m   g  .
 r



Figure 7.45
EXECUTE: (a) vB  2(980 m/s2 )(130 m)  160 m/s.
 (160 m/s)2

(b) n  (350 kg) 
 980 m/s 2   115  104 N.
 60 m

EVALUATE: The normal force n is the force that the tracks exert on the roller coaster car. The car exerts a force
of equal magnitude and opposite direction on the tracks.
7.47.(a) IDENTIFY: Use work-energy relation to find the kinetic energy of the wood as it enters the rough bottom.
SET UP: Let point 1 be where the piece of wood is released and point 2 be just before it enters the rough
bottom. Let y  0 be at point 2.
EXECUTE: U1  K2 gives K2  mgy1  784 J.
IDENTIFY: Now apply work-energy relation to the motion along the rough bottom.
SET UP: Let point 1 be where it enters the rough bottom and point 2 be where it stops.
K1  U1  Wother  K2  U 2
EXECUTE: Wother  W f  k mgs, K2  U1  U 2  0; K1  784 J
784 J  k mgs  0; solving for s gives s  200 m.
The wood stops after traveling 20.0 m along the rough bottom.
(b) Friction does 784 J of work.
EVALUATE: The piece of wood stops before it makes one trip across the rough bottom. The final mechanical
energy is zero. The negative friction work takes away all the mechanical energy initially in the system.
7.53.
IDENTIFY: Use the work-energy theorem, Eq. (7.7). Solve for K 2 and then for v2 .
SET UP: Let point 1 be at his initial position against the compressed spring and let point 2 be at the end of the
barrel, as shown in Figure 7.53. Use F  kx to find the amount the spring is initially compressed by the 4400 N
force.
K1  U1  Wother  K2  U 2
Take y  0 at his initial position.
EXECUTE: K1  0, K2  12 mv22
Wother  Wfric   fs
Wother  (40 N)(40 m)  160 J
Figure 7.53
U1,grav  0, U1,el  12 kd 2 , where d is the distance the spring is initially compressed.
F  kd so d 
F
4400 N

 400 m
k 1100 N/m
and U1,el  12 (1100 N/m)(400 m)2  8800 J
U 2,grav  mgy2  (60 kg)(980 m/s2 )(25 m)  1470 J, U 2 ,el  0
Then K1  U1  Wother  K2  U 2 gives
8800 J  160 J  12 mv22  1470 J
1 mv 2
2
2
 7170 J and v2 
2(7170 J)
 155 m/s
60 kg
EVALUATE: Some of the potential energy stored in the compressed spring is taken away by the work done by
friction. The rest goes partly into gravitational potential energy and partly into kinetic energy.
7.64.
IDENTIFY: Initially the ball has all kinetic energy, but at its highest point it has kinetic energy and potential
energy. Since it is thrown upward at an angle, its kinetic energy is not zero at its highest point.
SET UP: Apply conservation of energy: Kf  Uf  Ki  Ui . Let yi  0, so yf  h, the maximum height. At
this maximum height, vf , y  0 and vf ,x  vi,x , so vf  vi,x  (15 m/s)(cos60.0)  7.5 m/s. Substituting into
conservation of energy equation gives
EXECUTE: Solve for h: h 
1 mv 2
i
2
2
 mgh  12 m(75 m/s)2.
vi 2  (75 m/s)
(15 m/s)2  (75 m/s)2

 86 m
2g
2(980 m/s2 )
EVALUATE: If the ball were thrown straight up, its maximum height would be 11.5 m, since all of its kinetic
energy would be converted to potential energy. But in this case it reaches a lower height because it still retains
some kinetic energy at its highest point.
7.66.
IDENTIFY: Apply Eq. (7.14) to the initial and final positions of the truck.
SET UP: Let y  0 at the lowest point of the path of the truck. Wother is the work done by friction.
fr  r n  r mg cos  .
EXECUTE: Denote the distance the truck moves up the ramp by x. K1  12 mv02 , U1  mgL sin  , K2  0,
U 2  mgx sin  and Wother  - r mgx cos  . From Wother  ( K2  U 2 )  (K1  U1), and solving for x,
x
K1  mgL sin 
(v 2 /2 g )  L sin 
 0

mg  sin   r cos   sin   r cos 
EVALUATE: x increases when v0 increases and decreases when  r increases.
7.82.
IDENTIFY: Only gravity does work, so apply Eq. (7.4). Use F  ma to calculate the tension.
SET UP: Let y  0 at the bottom of the arc. Let point 1 be when the string makes a 45 angle with the vertical
and point 2 be where the string is vertical. The rock moves in an arc of a circle, so it has radial acceleration
arad  v2 /r
EXECUTE: (a) At the top of the swing, when the kinetic energy is zero, the potential energy (with respect to the
bottom of the circular arc) is mgl (1  cos ), where l is the length of the string and  is the angle the string
makes with the vertical. At the bottom of the swing, this potential energy has become kinetic energy, so
mgl (1  cos )  12 mv2 , or v  2 gl (1  cos )  2(980 m/s2 )(080m)(1  cos45)  21 m/s.
(b) At 45 from the vertical, the speed is zero, and there is no radial acceleration; the tension is equal to the
radial component of the weight, or mg cos  (012 kg)(980 m/s2 ) cos 45  083 N.
(c) At the bottom of the circle, the tension is the sum of the weight and the mass times the radial acceleration,
mg  mv22/l  mg (1  2(1  cos45))  19 N
EVALUATE: When the string passes through the vertical, the tension is greater than the weight because the
acceleration is upward.