PHYSICS 171
AQ 2009
Homework #7
1. Giancoli Chapter 8, Problem 8
The force is found from the relations on page 189.
U
U
Fx  
  6x  2 y 
Fy  
   2 x  8 yz 
x
y

F  ˆi  6 x  2 y   ˆj  2 x  8 yz   k 4 y 2
Fz  
U
z
 4 y 2

2. Giancoli Chapter 8, Problem 10
Use Eq. 8-6 to find the potential energy function.
U  x     F  x  dx  C    A sin  kx  dx  C 
A
k
cos  kx   C
3. Giancoli Chapter 8, Problem 16
(a) Since there are no dissipative forces present, the mechanical energy of the person–
trampoline–
Earth combination will be conserved. We take the level of the unstretched trampoline
as the zero level for both elastic and gravitational potential energy. Call up the positive
direction. Subscript 1 represents the jumper at the start of the jump, and subscript 2
represents the jumper upon arriving at the trampoline. There is no elastic potential
energy involved in this part of the problem. We have v1  4.5m s , y1  2.0 m, and
y2  0. Solve for v2 , the speed upon arriving at the trampoline.
E1  E2 
1
2
mv12  mgy1  12 mv22  mgy2 
v2   v12  2 gy1  
 4.5 m s 
2

 2 9.80 m s 2
1
2
mv12  mgy1  12 mv22  0 
  2.0 m    7.710 m s  7.7 m s
The speed is the absolute value of v2 .
(b) Now let subscript 3 represent the jumper at the maximum stretch of the trampoline, and
x represent the amount of stretch of the trampoline. We have v2  7.710 m s , y2  0,
x2  0, v3  0, and x3  y3. There is no elastic energy at position 2, but there is elastic
energy at position 3. Also, the gravitational potential energy at position 3 is negative,
and so y3  0. A quadratic relationship results from the conservation of energy
condition.
E2  E3 
1
2
1
2
mv22  mgy2  12 kx22  12 mv32  mgy3  12 kx32 
mv22  0  0  0  mgy3  12 ky32 
y3 


 mg  m 2 g 2  4  12 k   12 mv22


2  12 k 


  72 kg  9.80 m s 2 
ky32  mgy3  12 mv22  0 
1
2
 72 kg 
2
 mg  m 2 g 2  kmv22
k
 9.80 m s   5.8 10
 5.8 10 N m 
2
2
4

N m  72 kg  7.71m s 
4
 0.284 m , 0.260 m
Since y3  0 , y3  0.28 m . .
The first term under the quadratic is about 500 times smaller than the second term,
indicating that the problem could have been approximated by not even including
gravitational potential energy for the final position. If that approximation were made,
the result would have been found by taking the negative result from the following
solution.
E2  E3 
1
2
m
mv22  12 ky32  y3  v2
k
  7.71m s 
72 kg
5.8 104 N m
 0.27 m
4. Giancoli Chapter 8, Problem 20
Since there are no dissipative forces present, the mechanical energy of the roller coaster will be
conserved. Subscript 1 represents the coaster at point 1, etc. The height of point 2 is the
zero location for gravitational potential energy. We have v1  0 and y1  32 m.
Point 2:
1
2
mv12  mgy1  12 mv22  mgy2 ; y2  0  mgy1  12 mv22 

v2  2 gy1  2 9.80 m s2
Point 3:
1
2
  32 m  25m s
mv12  mgy1  12 mv32  mgy3 ; y3  26 m  mgy1  12 mv32  mgy3 

v3  2 g  y1  y3   2 9.80 m s2
Point 4:
1
2
  6 m   11m s
mv12  mgy1  12 mv42  mgy 4 ; y 4  14 m  mgy1  12 mv42  mgy1 


v4  2 g  y1  y4   2 9.80 m s2 18 m   19 m s
5. Giancoli Chapter 8, Problem 22
Draw a free-body diagram for each block. Write
Newton’s second law for each block. Notice that the acceleration of block A in the yA
is 0 zero.
 Fy1  FN  mA g cos  0  FN  mA g cos
F
x1
 FT  mA g sin   mAaxA
2
F
y2
 mB g  FT  mBa yB  FT  mB  g  a yB  Since the blocks are connected
by the cord, a yB  a xA  a. Substitute the expression for the tension force from the last
equation into the x direction equation for block 1, and solve for the acceleration.
mB  g  a   mA g sin   mA a  mB g  mA g sin   mA a  mBa
ag
m
B
 mA sin 
 mA  mB 

 9.80 m s  
2
5.0 kg  4.0 kg sin 32 
9.0 kg
 3.1m s 2
(b) Find the final speed of mB (which is also the final speed of mA ) using constant
acceleration relationships.
v 2f  v02  2a y  v 2f  2 g
v f  2 gh
m
B
 mA sin 
 mA  m B 

m
B
 mA sin 
 mA  m B 

2 9.80 m s 2
h

  0.75 m  
5.0 kg  4.0 kg sin 32 
9.0 kg
 2.2 m s
(c) Since there are no dissipative forces in the problem, the mechanical energy of the
system is conserved. Subscript 1 represents the blocks at the release point, and
subscript 2 represents the blocks when mB reaches the floor. The ground is the zero
location for gravitational potential energy for mB , and the starting location for mA is its
zero location for gravitational potential energy. Since mB falls a distance h, mA moves
a distance h along the plane, and so rises a distance h sin  . The starting speed is 0.
E1  E2  0  mA gh  12  mA  mB  v22  mB gh sin  
 mA  mB sin  

 mA  mB 
v2  2 gh 
This is the same expression found in part (b), and so gives the same numeric result.
6. Giancoli Chapter 8, Problem 23
At the release point the mass has both kinetic energy and elastic potential energy. The total
energy is 12 mv02  12 kx02 . If friction is to be ignored, then that total energy is constant.
(a) The mass has its maximum speed at a displacement of 0, and so only has kinetic energy
at that point.
k 2
x0
m
(b) The mass has a speed of 0 at its maximum stretch from equilibrium, and so only has
potential energy at that point.
1
2
2
mv02  12 kx02  12 mvmax
 vmax 
v02 
1
2
2
mv02  12 kx02  12 kxmax
 xmax 
x02 
m
k
v02
7. Giancoli Chapter 8, Problem 36
(a) Use conservation of energy to equate the potential energy at the top of the circular track to
the
kinetic energy at the bottom of the circular track. Take the bottom of the track to the be
0 level for gravitational potential energy.
2
Etop  Ebottom  mgr  12 mvbottom

vbottom 
2 gr 

2 9.80 m s 2
  2.0 m   6.261m s  6.3m s
(b)
The thermal energy produced is the opposite of the work done by the friction
force. In this
situation, the force of friction is the weight of the object times the coefficient of kinetic
friction.
Ethermal  Wfriction   Ffriction x   Ffriction x cos    k mg x  cos180   k mg x

  0.251.0 kg  9.80 m s 2
  3.0m   7.35J  7.4 J
(c) The work done by friction is the change in kinetic energy of the block as it moves from
point B to point C.

Wfriction  K  K C  K B  12 m vC2  vB2
vC 
2Wfriction
m
 vB2 
2  7.35J 
1.0 kg 


  6.261m s   4.9498 m s  4.9 m s
2
(d) Use conservation of energy to equate the kinetic energy when the block just contacts the
spring with the potential energy when the spring is fully compressed and the block has
no speed. There is no friction on the block while compressing the spring.
2
2
Einitial  Efinal  12 mvcontact
 12 kxmax

k m
2
vcontact
2
xmax
 4.9498 m s  2
 1.0 kg 
 0.20 m  2
 612.5 N m  610 N m
8. Giancoli Chapter 8, Problem 58
(a) The work to put m1 in place is 0, because it is still infinitely distant from the other two
masses.
Gm1m2
.
The work to put m2 in place is the potential energy of the 2-mass system, 
r12
The work to put m3 in place is the potential energy of the m1  m3 combination,

Gm1m3
, and the potential energy of the m2  m3 combination, 
r13
work is the sum of all of these potential energies, and so
Gm1m2 Gm1m3 Gm2 m3
W 



r12
r13
r23
 m1m2
W  G 

m1m3

Gm2 m3
r23
. The total
m2 m3 
 . Notice that the work is negative, which is a result
r23 
of the masses being gravitationally attracted towards each other.
(b) This formula gives the potential energy of the entire system. Potential energy does not
“belong” to a single object, but rather to the entire system of objects that interact to give
the potential energy.
 r12
r13
 m1m2 m1m3 m2 m3 


 is the binding energy of the system. It
r13
r23 
 r12
(c) Actually, W  G 
would take that much work (a positive quantity) to separate the masses infinitely far
from each other.
9. Giancoli Chapter 8, Problem 73
The net rate of work done is the power, which can be found by P  Fv  mav. The velocity is
dx
dv
 15.0t 2  16.0t  44 and a 
 30.0t  16.0.
given by v 
dt
dt
(a)


2

2
P  mav   0.28 kg  30.0  2.0   16.0 m s2 15.0  2.0   16.0  2.0   44  m s
 197.1W  2.0  102 W

(b) P  mav   0.28 kg  30.0  4.0   16.0 m s2 15.0  4.0   16.0  4.0   44  m s
 3844 W  3800 W
The average net power input is the work done divided by the elapsed time. The work done is
the change in kinetic energy. Note v  0   44 m s ,
v  2.0  15.0  2.0  16.0  2.0  44  16 m s , and
2
v  4.0  15.0  4.0  16.0  4.0  44  132 m s .
2
(c) Pavg

K
0 to 2.0
(d) Pavg
2.0 to 4.0

t
K
t


1
2

m v 2f  vi2
t
1
2

   0.28 kg   16 m s    44 m s   
m v 2f  vi2
t
2
1
2
2
2.0s
120 W
   0.28 kg  132 m s   16 m s    1200 W
2
1
2
2.0s
2