Download Chapter 6: Applications of Newton`s Laws

Chapter 6: Applications of Newton’s Laws
Solutions to Problems
43. Picture the Problem: Your car travels along a circular path at constant speed.
Strategy: Static friction between your tires and the road provides the centripetal force required to make the car travel
along a circular path. Set the static friction force equal to the centripetal force and calculate its value.
Solution: Set the static friction force equal to
the centripetal force:
mv 2  1200 kg   15 m/s 
f s  Fcp  macp 

 4.7 kN
r
57 m
2
2
Insight: The maximum static friction force is s mg   0.88   1200 kg   9.81 m/s   10.4 kN which corresponds to a
maximum cornering speed (without skidding) of 22 m/s.
49. Picture the Problem: Free-body diagrams for the top and bottom
positions on the Ferris wheel are drawn at right.
Strategy: Find the speed of the seat by dividing the circumference of its circular path by the time it takes to complete a cycle. Write Newton’s Second Law for the passenger at the top of the Ferris wheel and a second Newton’s Second Law for a passenger at the bottom, and solve each for the normal force.
Solution: 1. (a) The normal force on you by the Ferris wheel seat
is equal to your apparent weight. At the top of the Ferris wheel
your centripetal acceleration points downward and the normal
force is the smallest, but at the bottom of the Ferris wheel the
normal force must both support your weight and provide the
upward centripetal acceleration, so it is at its maximum there.
r
r
N
r
v
r
W
Bottom of Ferris Wheel
Top of Ferris Wheel
r
N
r
v
r
W
r
2. (b) Write Newton’s Second Law at the top of the
Ferris wheel and solve for your apparent weight N top :
F
3. Find the speed of the seat:
v
y
 N top  mg   macp   mv 2 r
N top  m  g  v 2 r 
2 r 2  7.2 m 

 1.6 m/s
T
28 s

 1.6 m/s  2
N top  m  g  v 2 r    55 kg   9.81 m/s 2 

7.2 m

4. Calculate the apparent weight N top :




 520 N  0.52 kN
5. Write Newton’s Second Law at the bottom of the
Ferris wheel and solve for your apparent weight N bottom :
F
 N bottom  mg  macp  mv 2 r
N bottom
 m  g  v r    55 kg   9.81 m/s 2 


y

2
 1.6 m/s  2 
7.2 m


 560 N  0.56 kN
Insight: The centripetal acceleration is 0.36 m/s so that the centripetal force is 3.6% of your 0.54 kN weight.
2
71. Picture the Problem: The free-body diagram of the airplane is depicted at right.
Strategy: Let the x axis point horizontally from the airplane towards the center of its
circular motion, and let the y axis point straight upward. Write Newton’s Second
Law in both the horizontal and vertical directions and use the resulting equations to
find θ and the tension T.
Solution: 1. (a) Write Newton’s
Second Law in the x and y
directions:
2. Solve the y equation for T and
substitute the result into the x
equation, and solve for θ:
T  mg cos 
 mg 
2
sin   m v r
 cos  
tan   v 2 rg
T sin   

  tan 1 


 1.21 m/s  2

2
 0.44 m   9.81 m/s  
 19
3. (c) Calculate the tension from
the equation in step 2:
Insight: This airplane is pretty small. The toy weighs only 0.74 N = 2.6 ounces and flies in a circle of diameter 2.9 ft.
92. Picture the Problem: The free-body diagram for the child in the seat is depicted at right.
Strategy: The swing accelerates towards the center of its circular path, so that the tension in
the string must provide an inward (horizontal) component of force on the child to keep him
traveling along the same circle. Find the centripetal acceleration from the speed and radius of
the path. Then write Newton’s Second Law in the vertical direction to find the tension in the
string as a function of θ. Finally, substitute the result into Newton’s Second Law for the
horizontal direction., and determine the angle θ.
θ
r
T
r
mg
Solution: 1. (a) Find the centripetal acceleration
from the speed and radius information:
2. (b) Write Newton’s Second Law in the y direction:
F
 T cos   mg  0  T  mg cos 
3. Write Newton’s Second Law in the x direction,
using equation 6-15 for the centripetal acceleration:
F
 T sin   macp  m v 2 r
4. Substitute the expression for T in step 1 into the
equation from step 2:
mg
mv 2
sin  
cos 
r
y
x
 tan  
v2
rg
  25 mi/h  0.447 m/s mi/h  2 
  tan 1 

 12 m   9.81 m/s2 

  47
5. (c) Because the weight and centripetal acceleration both depend linearly on the mass, the mass cancels out of the
expression for the angle θ. All swings are at the same angle, whether a child or an adult is present or even if the seat is
empty!
Insight: The riders are accelerating inward at about the same rate as the acceleration of gravity. They therefore
experience an outward “centrifugal” force equal to their weight.
Chapter 7: Work and Kinetic Energy
Answers to Even-Numbered Conceptual Exercises
8.
The appropriate ranking is C < A = D < B. To obtain this ranking, we use the fact that kinetic energy is proportional to mass to the first power, and proportional to speed to the second power. Thus, for example, jogger B has (1/2)(3)2 = 9/2 times the kinetic energy of jogger A.
Solutions to Problems
6.
Picture the Problem: The water skier is pulled horizontally.
Strategy: Multiply the force by the distance because in this case the two point along the same direction.
Solution: 1. (a) The work is positive because the force is along the direction of motion (θ = 0°).
2. (b) Apply equation 7-1 directly:
W  Fd   120 N   65 m   7800 J  7.8 kJ
Insight: While the work done by the rope is positive, the work done by friction is negative, so as long as the skier
moves at constant speed she doesn’t gain or lose kinetic energy.
7.
Picture the Problem: The water skier is pulled horizontally.
Strategy: Take the negative of the force times the distance because in this case the two point in opposite directions.
Solution: 1. (a) The work is negative because the force on the boat is opposite the direction of motion (θ = 180°).
2. (b) Apply equation 7-3 directly:
W  Fd cos    120 N   65 m  cos180o  7800 J  7.8 kJ
Insight: While the works done on the boat by the rope and by friction are negative, the work done by the motor is
positive, so as long as the boat moves at constant speed it doesn’t gain or lose kinetic energy.
10. Picture the Problem: The mop head is being pushed downward into the
floor.
Strategy: Use equation 7-3 keeping in mind the angle between the force
and the direction of motion.
F
θ
d
Solution: 1. (a) Use equation 7-3:
W  Fd cos    50.0 N   0.50 m  cos 55o  14 J
2. (b) If the angle is increased to 65°, a smaller component of the force will be along the direction of motion and
therefore the work done by the janitor will decrease.
Insight: Only the component of the force along the direction of the motion does any work. The vertical component of
the force increases the normal force.
18. Picture the Problem: The pine cone falls straight down under the influence of gravity.
Strategy: The work done by gravity equals the change in kinetic energy according to equation 7-7. The work done by
gravity is always W = mgh as indicated in Example 7-2 and Conceptual Checkpoint 7-1.
Solution: 1. (a) The work done by gravity on the pine
cone equals the increase in its kinetic energy. Set the
energies equal and solve for v:
W  K  mgh 
v
2 gh 
1
2
mv 2
2  9.81 m/s 2   16 m   18 m/s
2. (b) Air resistance did negative work because the speed and therefore the kinetic energy of the pine cone when it
landed was reduced. Air resistance removed energy from the pine cone.
Insight: Kinetic friction always does negative work because the force is always opposite to the direction of motion.
34. Picture the Problem: The runner does work against gravity as he elevates his center of gravity.
Strategy: The power required is the work required to change the elevation divided by the time. As in Conceptual
Checkpoint 7-1 and Example 7-2 the work required to change the elevation of an object is W  mgh.
Solution: Divide the work required by the time:
2
W mgh  70.0 kg   9.81 m/s   1576 steps  0.20 m/step 
P


t
t
 9 min  60 s/min  33 s 
P  380 W  1 hp 746 W  0.51 hp
Insight: The energy and power required of the runner is much higher than this because muscles aren’t 100% efficient at
converting food energy into mechanical energy, and the body requires additional energy to stay warm, keep the heart
pumping, etc.
44. Picture the Problem: The car accelerates horizontally in a straight line.
Strategy: The power required is the work required to change the kinetic energy divided by the time. Use ratios to
easily find the desired quantities.
Solution: 1. (a) Combine equations 7-7 and 7-10:
2. Now divide top and bottom by
for the velocities:
1
2
m and substitute
t
2
2
W K 12 mvf  12 mvi

 1 2
P
P
mvi T
2
t
2
vf2  vi2  2v   v

 3T
vi2 T
v2 T
2

3. (b) Again combine equations 7-7 and 7-10:
4. Now solve for vf :
1
2
1
2
W  K  12 mvf2  Pt  

mv 2 
 2T 
T 
mvf2  mv 2 so vf2  2v 2 and thus vf  v 2
Insight: The assumption that the power remains constant is not realistic because car engines only generate their rated
horsepower at high engine rpm, so less power is generated when the car first begins to accelerate.
45. Picture the Problem: The protein exerts a force that is parallel to the displacement.
Strategy: The work done by the protein equals the force times the displacement since the two vectors are parallel.
Solution: Apply equation 7-1 directly:
W  Fd   7.5  10 12 N   8.0  10 9 m   6.0  1020 J
Insight: A convenient unit of energy at this scale is the electron volt (eV), where 1.0 eV = 1.60×10-19 J. The work done
in this problem by the protein is thus 0.375 eV.
53. Picture the Problem: The glove rises vertically to height h, stops, and returns to the ground.
Strategy: Gravity does negative work on the glove, reducing its kinetic energy to zero when it reaches height h. Set the
work done by gravity equal to the change in kinetic energy between the ground and h/2.
Solution: 1. Combine equations 7-1 and 7-7:
W  Fd    mg   h 2   K f  K i
K f  K i  mgh 2
2. Now set the work done by gravity when the
glove rises to height h equal to the initial kinetic
W  Fd    mg  h  K f  K i  0  K i
K i  mgh
energy:
3. Substitute the result into the first equation:
K f  K i  mgh 2  mgh  mgh 2  mgh 2  K i 2  K 2
Insight: The work done by gravity converts the glove’s kinetic energy into gravitational potential energy, a topic
discussed in Chapter 8.
70. Picture the Problem: The block slides horizontally but the pulling force is
inclined at an angle.
Strategy: Use the definition of work to find the angle between the force and
displacement vectors. Then set the work done by the pulling force equal to
the change in kinetic energy of the block.


50.0 J
 W 
1
o
  42.2
  cos 
Fd
45.0
N
1.50
m

 


 
Solution: 1. (a) Solve equation 7-3 for θ:
W  Fd cos  so   cos 1 
2. (b) Solve equation 7-7 for m:
W  K  12 mvf2  0 so m 
2  50.0 J 
2W

 14.8 kg
vf2
 2.60 m/s  2
Insight: Only the component of the force along the direction of the motion does any work. The vertical component of
the force reduces the normal force a little.