Dynamics Extra Study Questions
Short Answer
1. An object with a mass of 15 kg rests on a frictionless horizontal plane and is acted upon by a horizontal force
of 30 N.
(a) What is its acceleration?
(b) How far will it move in 10 s?
(c) What will be its velocity after 10 s?
2. (a) What is the net force required to give an automobile of mass 1600 kg an acceleration of 4.5 m/s2?
(b) What is the acceleration of a wagon of mass 20 kg if a horizontal force of 64 N is applied to it (ignore
friction)?
(c) What is the mass of a block of iron if a net force of 240 N causes it to accelerate across a smooth
horizontal surface at 2.5 m/s2?
3. A 1.0 kg toy car is moving across a smooth floor with a velocity of 5.0 m/s. An unbalanced force of 2.0 N
acts on the car for 4.0 s. Determine the velocity of the car at the end of the interval in each of the following
cases.
(a) if the force acts in the direction of motion of the car
(b) if the force acts in the opposite direction to the motion of the car
4. State the reaction force for each of the following forces.
(a) the southward force of a field goal kicker’s toe on a football
(b) the backward force of a jogger’s shoe on the ground
(c) the downward force of a book on a desk
(d) the backward force of a jet’s engines on its exhaust gases
(e) the backward pull of a swimmer’s hands on the water in the butterfly stroke
5. A beginning physics student, confused by a seeming contradiction in Newton’s laws, asks her teacher the
following question: “If, for every force there is an equal and opposite reaction force, then all forces in nature
come in equal and opposite pairs, and are therefore balanced. Thus, since there can never be such a thing as an
unbalanced force, how can any object ever accelerate?” Explain the fault in this common misconception.
6. A fireman at the scene of a fire is holding a heavy hose out of which water is gushing. To keep his balance, he
often has to lean. Which way does he lean, forward or backward, and why?
7. A squirrel with an armful of nuts is sliding helplessly across the flat, icy roof, getting dangerously close to the
edge. He understands Newton’s Third Law, and is able to save himself. Explain how he does it.
8. A rocket carrying the Space Shuttle blasts off from the Kennedy Space Center. In which of the following
positions will it have a greater acceleration, assuming that the engines exert a constant force?
(a) at ground level
(b) 50 m above ground level
Explain your answer.
9. Two crates, of mass 12.0 kg and 20.0 kg, respectively, are pushed across a smooth floor together, the 20 kg
crate in front of the 12 kg crate. Their acceleration is 1.75 m/s2. Calculate each of the following.
(a) the force applied to push the crates
(b) the action-reaction forces between the two crates
Recalculate (a) and (b) if the relative positions of the two crates are reversed.
10. A locomotive with a mass of 2.0 × 104 kg accelerates from rest to a velocity of 2.0 m/s in 5.0 s. If it is pulling
a train of 20 cars, each of mass 1.0 × 104 kg, what is the force in the coupling at each of the following points?
(a) between the locomotive and the first car
(b) between the 10th car and the 11th car
11. Three small children of mass 20.0 kg, 24.0 kg, and 16.0 kg, respectively, hold hands, as shown, and are pulled
across a smooth frozen pond by a larger boy on skates, who pulls a horizontal rope being held by the first
child. The skater pulls on the rope with a force of 135 N. Calculate each of the following.
(a) the acceleration of the skater
(b) the force with which each pair of children must hold hands, to ensure that the chain is not broken
12. A 3.0 kg toy is pulled by a force of 24 N. If the toy starts from rest, how far will it travel in the first 5.0 s?
13. A 40 kg sprinter starts from rest and 2.0 s later is running at a speed of 8.0 m/s. What is the average net
horizontal force acting on her? What exerts this force?
14. An 8.0 g bullet travelling at 400 m/s passes through a heavy block of wood in 4.0 × 10–4 s, emerging with a
velocity of 100 m/s. Ignore any motion of the wood.
(a) With what average force did the wood oppose the motion of the bullet?
(b) How thick is the block of wood?
15. A 0.22 calibre rifle shoots a bullet of mass 1.8 g with a muzzle velocity of 500 m/s. If the barrel is 25 cm
long, what is the average force exerted on the bullet while it is in the barrel?
16. A sled of 6.0 kg mass is moving along a smooth, horizontal ice surface with a velocity of v0. A force of 36 N
is applied to the sled in its direction of motion, increasing its velocity to 2v0 while it moves 10 m. Find the
following.
(a) the sled’s original velocity, v0
(b) the length of time that the force acted.
17. A gardener pushes down along the handle of a lawn mower of 20 kg mass with a force of 150 N. The handle
makes an angle of 60° with the ground. Calculate the instantaneous acceleration of the mower if the frictional
force between its wheels and the ground at that instant is 25 N.
18. A man drags a package across the floor with a force of a 40 N, as shown. The mass of the package is 10 kg. If
the acceleration of the package is 3.5 m/s2, and friction can be neglected, at what angle to the horizontal does
the man pull?
19. A boy with a mass of 30 kg pulls a cart with a mass of 100 kg towards himself by a rope. With what force
does he have to pull on the rope to accelerate the cart at 2.0 m/s2? With what force must his feet push on the
ground to keep him from moving towards the cart? If there is no friction between his feet and the ground,
what is his acceleration?
20. A cart is pulled in each of several trials, with a different number of stretched elastic bands. A constant
acceleration is observed in each trial. The graph shows acceleration versus force exerted by the stretched
bands.
(a) What is the mass of the cart?
(b) An extrapolation of the graph does not pass through the origin. What does this indicate?
21. A car with a mass of 1000 kg is moving in a straight line at a constant speed of 30 m/s. It is brought to rest in
25 s. What constant force is acting to stop the car?
22. An electron has a mass of 9.1 × 10–31 kg. Between the electrodes of a cathode-ray tube, it moves a distance of
4.0 mm, accelerated by a net electrical force of 5.6 × 10–15 N. Assuming that it started from rest, find its
acceleration and its final velocity.
23. A speedboat is pulling two water skiers, using two ropes attached to the back of the boat. Each rope makes an
angle of 30° on either side of the boat’s axis. The force exerted by each rope is 400 N. If the boat is moving in
a straight line at constant speed, what force must it be exerting to keep the pair of skiers moving?
24. A block of mass 2.0 kg is placed on a smooth plane, inclined to the horizontal at an angle of 15°. The force of
gravity acting straight down on the block is 20 N.
(a) What is the acceleration of the block down the plane?
(b) How far up the plane was the block released, if it took 1.5 s to reach the bottom after it was released from
rest?
25. What is the force of gravity acting on a 1000 kg car resting on Earth?
26. What is the force of gravity at Earth’s surface on
(a) a 50 kg girl, and
(b) a 100 g bullet?
27. If the gravitational field constant on the surface of the moon is 1.6 N/kg, what is the force of gravity on the
following if they are on the surface of the moon?
(a) a 50 kg girl, and
(b) a 100 g bullet?
28. A horizontal force of 50 N is required to pull an 8.0 kg block of aluminum at a uniform velocity across a
horizontal wooden desk. What is the coefficient of kinetic friction?
29. The driver of a 2000 kg car applies the brakes on a dry concrete roadway. Calculate the force of friction
between the tires and the road surface. Note: the coefficient of kinetic friction (µk) on dry concrete is 1.02.
30. It takes a 50 N horizontal force to pull a 20 kg object along the ground at a constant velocity. What is the
coefficient of friction?
31. If the coefficient of friction is 0.30, how much horizontal force is needed to pull a mass of 15 kg across a level
board at a uniform velocity?
32. The gravitational field constant at the Moon’s surface is 1.6 N/kg. A moon rock is dropped from a height of
10 m. How long will it take to reach the surface of the moon?
33. The gravitational field constant 150 km above the Earth’s surface is 9.5 N/kg. With this in mind, what is
(a) the force of gravity on a 60 kg astronaut at this altitude
(b) what will be the final velocity of a rock if it falls freely at this altitude for 10 s, starting at rest?
34. A rock is thrown horizontally at 10.0 m/s from the top of a cliff 122.5 mm high.
(a) How long does the rock take to reach the ground?
(b) What is the horizontal displacement of the rock?
35. A ball is thrown from the top of a building with a horizontal velocity of 20 m/s. It hits level ground 80 m from
the face of the building. How high is the building?
36. If a baseball player throws a ball at 35.0 m/s, what is its maximum range?
37. A player kicks a soccer ball towards the goalkeeper, but at an angle of 37° to the horizontal and with an initial
speed of 14.7 m/s. The goalkeeper stands 26.0 m from the kicker. Where will the ball land relative to the
goalkeeper?
38. A 2.00 kg stone is whirled in a circle by a rope 4.00 m long, completing five revolutions in 2.00 s. Calculate
the tension in the rope if the stone is rotated horizontally on a smooth frictionless surface.
39. A stone of mass 1.00 kg is attached to one end of a string 1.00 m long, of breaking strength 500 N, and is
whirled in a horizontal circle on a frictionless table top. The other end of the string is kept fixed. Find the
maximum speed the stone can attain without breaking the string.
40. A gun shoots a bullet at 1200 m/s at an angle of 60° above the horizontal. Neglecting air resistance, determine
(a) its time of flight
(b) its range
41. A ball projected horizontally from a ceiling height of 3.0 m hits the floor 4.0 m “down range.” Calculate the
following.
(a) its time of flight
(b) its horizontal velocity
42. If you can hurl a ball so that its initial speed is 30 m/s, what is the widest river you can throw it across?
43. A rifle with a muzzle velocity of 460 m/s shoots a bullet at a small target 800 m away at the same height. At
what angle above the horizontal must the gun be aimed so that the bullet will hit the target?
44. You have determined the following results when doing an investigation. Using proportioning techniques, find
the new value for the centripetal force.
Before
mass = 1 ball
radius = 0.75 m
frequency = 1.5 Hz
centripetal force = 8.0 units
After
mass = 3 balls
radius = 1.50 m
frequency = 3.0 Hz
centripetal force = ? units
45. A 200 g ball on the end of a string is rotated in a horizontal circle of radius 10.0 m. The ball completes 10
rotations in 5.0 s. What is the centripetal force of the string on the ball?
Problem
46. A 0.50 kg skateboard is at rest on a rough, level floor on which two lines have been drawn 1.0 m apart. A
constant horizontal force is applied to the skateboard at the beginning of the interval, and is removed at the
end. The skateboard takes 8.5 s to travel the 1.0 m distance, and it then coasts for another 1.25 m before
coming to rest. Calculate the force applied to the skateboard, and also the constant frictional force opposing
its motion.
47. A small boy pulls his wagon, of mass 24 kg, giving it a horizontal acceleration of 1.5 m/s2. If the wagon’s
handle makes an angle of 40° with the ground while the boy is pulling on it, and there is a frictional force of
6.0 N opposing the wagon’s motion, with what force is he pulling on the handle of the wagon?
48. Forces of 100 N [N] and 80 N [W] act simultaneously on an object of mass 10 kg. What is the acceleration of
the object?
49. Forces of 2.0 N and 1.0 N act on an object of mass 5.0 kg, as shown in the diagram.
(a) Calculate the net force acting on the object.
(b) What is the acceleration of the object?
50. An empty railway flatcar of mass 15 000 kg is being pulled along a smooth, horizontal track by a tractor
travelling on a road parallel to the track. The rope joining the tractor and the flatcar makes an angle of 25°
with the track.
(a) If the acceleration of the flatcar is 0.80 m/s2, what is the force exerted by the rope on the flatcar?
(b) Why does the flatcar have no sideways acceleration? A sideways force is exerted by the rope!
51. Two girls, one of mass 40 kg and the other of mass 60 kg, are standing side by side in the middle of a frozen
pond. One pushes the other with a force of 360 N for 0.10 s. The ice is essentially frictionless.
(a) What is each girl’s acceleration?
(b) What velocity will each girl acquire in the 0.10 s that the force is acting?
(c) How far will each girl move during the same time period?
52. A motorist has a reaction time of 0.60 s. (Reaction time is the interval between seeing a danger and applying
the brakes.) While driving at 72 km/h, he sees a child run suddenly onto the road, 40 m in front of his car. If
the mass of the car is 1000 kg and the average horizontal force supplied during braking is 8000 N, will he be
able to stop in time to avoid hitting the child?
53. A child’s wagon experiences a frictional force of 73 N whenever it is in motion, regardless of the load it is
carrying. An applied horizontal force of 128 N causes the wagon to accelerate at 5.0 m/s2. The same applied
force, with a child on the wagon, causes it to accelerate at 1.0 m/s2. What is the mass of the child?
54. A net force of 8.0 N gives a mass m1 and acceleration of 2.0 m/s2 and a mass m2, an acceleration of 4.0 m/s2.
What acceleration would the force give the two masses if they were fastened together?
55. The graph shows the velocity of a 5.0 kg radio-controlled toy car, moving in a straight line, as a function of
time. Plot a force-time graph for the car.
56. Two girls pull a sled across a field of snow, as shown in the diagram. A third girl pulls backward with a 2.0 N
force. If the mass of the sled is 10 kg, determine its instantaneous acceleration.
57. A plane takes off from a level runway with two gliders in tow, one behind the other. The first glider has a
mass of 1600 kg and the second a mass of 800 kg. The frictional drag may be assumed as constant and equal
to 2000 N on each glider. The towrope between the first glider and the plane can withstand a tension of 10
000 N.
(a) If a velocity of 40 m/s is required for takeoff, how long a runway is needed?
(b) How strong must the towrope between the two gliders be?
58. A baby carriage with a mass of 50 kg is being pushed along a rough sidewalk with an applied horizontal force
of 200 N, and it has a constant velocity of 3.0 m/s.
(a) What other horizontal force is acting on the carriage, and what is the magnitude of that force?
(b) What value of applied horizontal force would be required to accelerate the carriage from rest to 7.0 m/s in
2.0 s?
59. A bullet of mass 20 g strikes a fixed block of wood at a speed of 320 m/s. The bullet embeds itself in the
block of wood, penetrating to a depth of 6.0 cm. Calculate the average net force acting on the bullet while it is
being brought to rest.
60. A passenger in an elevator has a mass of 100 kg. Calculate the force, in newtons, exerted on the passenger by
the elevator, if the elevator is
(a) at rest
(b) moving with an upward acceleration of 30 cm/s2
(c) moving with a downward acceleration of 15 cm/s2
(d) moving upward with a uniform velocity of 15 cm/s
(e) falling freely (the cable breaks).
61. An elevator, complete with contents, has a mass of 2000 kg. By drawing free-body diagrams and by
performing the necessary calculations, determine the value of T (the tension in the elevator cable) in the
following:
(a) when the elevator is at rest
(b) when the elevator is moving toward at a constant velocity of 2.0 m/s
(c) when the elevator is moving downward at a constant velocity of 2.0 m/s
(d) when the elevator is accelerating upward at 1.0 m/s2
(e) when the elevator is accelerating downward at 1.0 m/s2
62. A man measures the acceleration of an elevator by using a spring balance. He fastens the scale to the roof, and
suspends a mass from it. If the scale reads 98 N when the elevator is at rest, and 93 N when the elevator is
moving, find the following.
(a) the acceleration of the elevator
(b) the direction in which the elevator is accelerating
63. A 2.0 kg mass, placed on a smooth, level table, is attached by a light string passing over a frictionless pulley
to a 5.0 kg mass hanging freely over the edge of the table, as illustrated. Calculate
(a) the tension in the string
(b) the acceleration of the 2.0 kg mass.
64. Two spheres of masses 1.5 kg and 3.0 kg are tied together by a light string looped over a frictionless pulley.
They are allowed to hang freely. What will be the acceleration of each mass? Assume that up is positive and
down is negative.
65. A 5.0 kg mass rests on a level, frictionless table, attached to a 3.0 kg mass by a light string that passes over a
frictionless pulley. Calculate the tension in the string when the masses are released.
66. A 40 kg block on a level, frictionless table is connected to a 15 kg mass by a rope passing over a frictionless
pulley. What will be the acceleration of the 15 kg mass when it is released?
67. A 3.0 kg mass is attached to a 5.0 kg mass by a strong string that passes over a frictionless pulley. When the
masses are allowed to hang freely, what will be
(a) the acceleration of the masses
(b) the magnitude of the tension in the string
68. A 20 kg box is dragged across a level floor with a force of 100 N. The force is applied at an angle of 40°
above the horizontal. If the coefficient of kinetic friction is 0.32, what is the acceleration of the box?
69. A boy on a toboggan is sliding down a snow-covered hillside. The boy and toboggan together have a mass of
50 kg, and the slope is at an angle of 30° to the horizontal. Find the boy’s acceleration considering the
following.
(a) if there is no friction
(b) if the coefficient of kinetic friction is 0.15
70. A cart with a mass of 2.0 kg is pulled across a level desk by a horizontal force of 4.0 N. If the coefficient of
kinetic friction is 0.12, what is the acceleration of the cart?
71. A girl pushes a light snow shovel at a uniform velocity across a sidewalk. If the handle of the shovel is
inclined at 55° to the horizontal and she pushes along the handle with a force of 100 N, what is the force of
friction? What is the coefficient of kinetic friction?
72. A 10 kg block of ice slides down a ramp 20 m long, inclined at 10° to the horizontal.
(a) If the ramp is frictionless, what is the acceleration of the block of ice?
(b) If the coefficient of kinetic friction is 0.10, how long will it take the block to slide down the ramp, if it
starts from rest?
Note: The kinetic friction is implied.
73. A skier has just begun descending a 20° slope. Assuming that the coefficient of kinetic friction is 0.10,
calculate
(a) the acceleration of the skier
(b) his final velocity after 8.0 s
74. The graph above shows the relationship between the force of air resistance and the velocity of a falling 1.0 kg
steel ball. What is the instantaneous acceleration of the ball when it is travelling downward at 10.3 m/s?
75. Using the graph above, answer the following questions.
(a) What is the acceleration when the ball is falling at 8.0 m/s?
(b) What is the terminal velocity of the steel ball?
(c) If the steel ball is fired vertically downward at 22 m/s, what will be its initial acceleration?
76. A helicopter is rising vertically at a uniform velocity of 14.7 m/s. When it is 196 m from the ground, a ball is
projected from it with a horizontal velocity of 8.5 m/s with respect to the helicopter. Calculate the following.
(a) when the ball will reach the ground
(b) where it will hit the ground
(c) what its velocity will be when it hits the ground
77. A stone thrown horizontally from the top of a tall building takes 7.56 s to reach the street. How high is the
building?
78. A bullet is projected horizontally at 300 m/s from a height of 1.5 m. Ignoring air resistance, calculate how far
it travels horizontally before it hits the ground. (Assume that the ground is level.)
79. An object is projected horizontally with a velocity of 30 m/s. It takes 4.0 s to reach the ground. Neglecting air
resistance, determine the following.
(a) the height at which the object was projected
(b) the magnitude of the resultant velocity, just before the object strikes the ground
80. A bomber in level flight, flying at 92.0 m/s, releases a bomb at a height of 1950 m.
(a) How long is it before the bomb strikes Earth?
(b) How far does it travel horizontally?
(c) What are the horizontal and vertical components of its velocity when it strikes?
(d) What is its velocity of impact?
81. A cannonball shot horizontally from the top of a cliff with an initial velocity of 425 m/s is aimed towards a
schooner on the ocean below. If the cliff is 78 m above the ocean surface, calculate the following:
(a) the time for the cannonball to reach the water
(b) the horizontal displacement of the cannonball
(c) the velocity of the cannonball just before it strikes the water
82. A balloon is rising at a vertical velocity of 4.9 m/s. At the same time, it is drifting horizontally with a velocity
of 1.6 m/s. If a bottle is released from the balloon when it is 9.8 m above the ground, determine the following.
(a) the time it takes for the bottle to reach the ground
(b) the horizontal displacement of the bottle from the balloon
83. A cannonball is fired with a velocity of 100 m/s at 25° above the horizontal. Determine how far away it lands
on level ground.
84. A 3.5 kg steel ball is swung at a constant speed in a vertical circle of radius 1.2 m, on the end of a light, rigid
steel rod, as illustrated. If the ball has a frequency of 1.0 Hz, calculate the tension in the rod due to the mass at
the top (A) and at the bottom (B) positions.
85. The pilot of an airplane, which has been diving at a speed of 540 km/h, pulls out of the dive at constant speed.
(a) What is the minimum radius of the plane’s circular path in order that the acceleration of the pilot at the
lowest point will not exceed 7g?
(b) What force is applied on an 80 kg pilot by the plane seat at the lowest point of the pull-out?
86.
A horizontal force is applied to a 2.0 kg block, moving on a level table. A force that is one-quarter the force
of gravity on the block is required to move it at a constant velocity. Calculate the force necessary to accelerate
the moving block from rest to a speed of 3.0 m/s in 4.0 s.
87. A space traveller has landed on the surface of an unknown planet similar to Earth. He drops a small lead ball
from the top of his space ship and finds it takes 3.0 s to reach the ground, 18 m below. If the force of gravity
on the astronaut is 710 N on Earth, how much will it be on the planet?
88. An aerospace scientist has designed a rocket with a mass of 1.0 × 103 kg. He wants it to accelerate straight up
with an initial acceleration of 21 m/s2. What thrust must the rocket engine develop?
89. A rocket of mass 1.0 × 103 kg is being fired to a height of 5.0 × 103 m. The rocket engine shuts off when the
rocket reaches a height of 1.0 × 103 m, and the rocket coasts up to 5.0 × 103 m.
(a) Draw a free-body diagram to show the forces acting on the rocket (i) while the engine is on and (ii) after
the engine shuts off.
(b) What velocity must the rocket have at the 1.0 × 103 m point to enable it to reach 5.0 × 103 m?
(c) What acceleration did the rocket experience while the engine was on? off?
(d) What force did the rocket engine exert on the rocket?
90. An exceptional vertical jump from rest would raise a person 0.80 m off the ground. To do this, what constant
force would a 70.0 kg person have to exert against the ground? Assume the person lowers himself by 0.20 m
prior to jumping and remains in a standing position while in the air.
91. A 5000 kg helicopter accelerates upward at 0.50 m/s2 while lifting a 2000 kg car.
(a) What is the lift force exerted by the air on the rotors?
(b) What is the tension in the cable that connects the car to the helicopter?
92. Jane wishes to quickly scale a slender vine to visit Tarzan in his treetop hut. The vine is known to safely
support the combined weight of Tarzan, Jane, and Cheetah. Tarzan has twice the mass of Jane, who has twice
the mass of Cheetah. If the vine is 60 m long, what minimum time should Jane allow for the climb?
93. A boy pushing a 20 kg lawn mower exerts a force of 100 N along the handle. If the handle is elevated 37° to
the horizontal, determine
(a) the component of the applied force that pushes the lawn mower forward
(b) the acceleration of the lawn mower, if the frictional force is 60 N
(c) the component of the applied force that pushes the lawn mower vertically toward the ground
(d) the gravitational force exerted on the mower
(e) the total downward force of the mower on the ground, when pushed
(f) the normal force exerted on the mower by the ground
(g) the effective coefficient of kinetic friction.
94. An 80 kg man is standing in an elevator on a set of spring scales calibrated in newtons. Suppose the elevator
accelerates downward at 3.0 m/s2. What reading will the scales have?
95. An empty elevator of mass 2.7 × 103 kg is pulled upward by a cable with an upward acceleration of 1.2 m/s2.
(a) What is the tension in the cable?
(b) What would the tension be if the elevator were accelerating downward at 1.2 m/s2?
(c) Does the direction of motion of the elevator matter in (a) and (b)?
96. A fish hangs from a spring scale supported from the roof of an elevator.
(a) If the elevator has an upward acceleration of 1.2 m/s2 and the scale reads 200 N, what is the true force of
gravity on the fish?
(b) Under what circumstances will the scale read 150 N?
(c) What will the scale read if the elevator cable breaks?
97. For each of the following systems:
(i) draw a free-body diagram for each mass.
(ii) find the tension in the string(s),
(iii) find the rate at which the masses will accelerate, and
(iv) calculate how far each mass will move in 1.2 s, if the system starts from rest.
(Assume that both the surfaces and the pulleys are frictionless.)
(a)
(b)
(c)
98. Two masses are connected by a light cord over a frictionless pulley, as illustrated. The coefficient of kinetic
friction is 0.18.
(a) What is the acceleration of the system?
(b) What is the tension in the cord?
(c) Assume that the system starts to move. For what values of will it not start?
99. What is the acceleration of the system illustrated, if the coefficient of kinetic friction is 0.20? Assume that it
starts.
100. Tarzan (mass 100 kg) holds one end of an ideal vine (infinitely strong, completely flexible, but having zero
mass). The vine runs horizontally to the edge of a cliff, then vertically to where Jane (mass 50 kg) is hanging
on, above a river filled with hungry crocodiles. A sudden sleet storm has removed all friction. Assuming that
Tarzan hangs on, what is his acceleration towards the cliff edge?
101. An engineer designs a jet boat with a mass of 6.0 × 102 kg. The force of water resistance is given by Fr = 1.8
v2. (Fr is the resistance in N, and v is the speed in m/s.)
(a) What thrust (force) must the engine develop to enable the boat to reach a speed of 40 m/s?
(b) What is the maximum acceleration the boat can achieve with this thrust?
102. A 70 kg hockey player coasts along the ice on steel skates. If the coefficient of kinetic friction is 0.010.
(a) What is the force of friction?
(b) How long will it take him to coast to a stop, if he is travelling at 1.0 m/s?
103. A small 10 kg cardboard box is thrown across a level floor. It slides a distance of 6.0 m, stopping in 2.2 s.
Determine the coefficient of friction between the box and the floor.
104. A 0.5 kg wooden block is placed on top of a 1.0 kg wooden block. The coefficient of static friction between
the two blocks is 0.35. The coefficient of kinetic friction between the lower block and the level table is 0.20.
What is the maximum horizontal force that can be applied to the lower block without the upper block
slipping?
105. A boy pulls a 50 kg crate across a level floor with a force of 200 N. If the force acts at an angle of 30° up
from the horizontal, and the coefficient of kinetic friction is 0.30, determine the following.
(a) the normal force exerted on the crate by the floor
(b) the horizontal frictional force exerted on the crate by the floor
(c) the acceleration of the crate
106. A can of pop is given a shove. It slides across a table, eventually coming to a stop. If its initial velocity is 2.0
m/s, and the coefficient of kinetic friction between the two surfaces is 0.20, how far will it travel across the
table?
107. A skier skiing downhill reaches the bottom of a hollow with a velocity of 20 m/s, and then coasts up a hill
with a 10° slope. If the coefficient of kinetic friction is 0.10, how far up the slope will she travel before she
stops?
108. A multi-flash, stroboscopic photograph is taken of a freely-falling polystyrene ball. The flash rate of the
stroboscope is 20 Hz. The resulting photograph is one-fifth actual size.
(a) What is the time interval between pictures?
(b) Calculate the average speed of the ball between each flash, in metres per second.
(c) Calculate (i) the increase in speed between successive ball positions, and (ii) the acceleration between
successive ball positions.
(d) Explain why the acceleration appears to decrease as the ball falls
109. What force must a track star exert on a shot put of mass 5.0 kg to accelerate it from rest to a velocity of 6.0
m/s [up] while pushing it through a vertical distance of 80 cm? How long will it rise after leaving the shotputter’s hand?
110. A robber jumps from an apartment window 5.0 m high. When he strikes the ground below, he bends his knees
so that his torso decelerates over an approximate distance of 0.70 m. If the mass of his torso (excluding his
legs) is 50 kg, find the following.
(a) his velocity just before his feet strike the ground
(b) the force exerted on his torso by his legs during the deceleration
111. A ball is thrown horizontally from a window at 10 m/s and hits the ground 5.0 s later. What is the height of
the window and how far from the base of the building does the ball hit?
112. A cannon is fired at 30° above the horizontal with a velocity of 200 m/s from the edge of a cliff 125 m high.
Calculate where the cannonball lands on the level plain below.
113. A shell is fired horizontally from a powerful gun, located 44 mm above a horizontal plane, with a muzzle
speed of 245 m/s.
(a) How long does the shell remain in the air?
(b) What is its range?
(c) What is the magnitude of the vertical component of its velocity as it strikes the target?
114. A bomber, diving at an angle of 53° with the vertical, releases a bomb at an altitude of 730 m. The bomb hits
the ground 5.0 s after being released.
(a) What was the velocity of the bomber?
(b) How far did the bomb travel horizontally during its flight?
(c) What were the horizontal and vertical components of its velocity just before striking the ground?
115. A driver, accelerating too quickly on a horizontal bridge, skids, crashes through the bridge railing, and lands
in the river 20.0 m below the level of the bridge roadway. The police find that the car is not vertically below
the break in the railing, but is 53.6 m beyond it horizontally.
(a) Determine the speed of the car before the crash, in km/h.
(b) What properties of falling bodies did you assume in making your calculation in (a)?
(c) State whether your answer in (a) is an overestimate or an underestimate, and why.
116. An artillery gun is fired so that its shell has a vertical component of a velocity of 210 m/s and a horizontal
component of 360 m/s. If the target is at the same level as the gun, and air friction is neglected,
(a) how long will the shell stay in the air?
(b) how far down-range will the shell hit the target?
117. A baseball, thrown from shortstop position to first base, travels 32 m horizontally, rises 3.0 mm, and falls 3.0
m. Find the initial velocity of the ball.
118. A player kicks a football with an initial velocity of 15 m/s at an angle of 42° above the horizontal. A second
player standing at a distance of 30 m from the first, in the direction of the kick, starts running to meet the ball
at the instant it is kicked. How fast must he run in order to catch the ball before it hits the ground?
119. In the Bohr model of the hydrogen atom, the electron revolves around the nucleus. If the radius of the orbit is
5.3 × 10–11 m and the electron makes 6.6 × 1015 r/s, find the following.
(a) the acceleration of the electron.
(b) the centripetal force acting on the electron (This force is due to the attraction between the positively
charged nucleus and the negatively charged electron.) The mass of the electron is 9.1 × 10–31 kg.
120. A string pendulum 1.12 m long has a bob with a mass of 200 g.
(a) What is the tension in the string when the pendulum is at rest?
(b) What is the tension at the bottom of the swing, if the pendulum is moving at 1.2 m/s?
121. When you whirl a ball on a cord in a vertical circle, you find a critical speed at the top for which the tension in
the cord is zero. This is because the force of gravity on the object itself supplies the necessary centripetal
force. How slowly can you swing a 2.5 kg ball like this so that it will just follow a circle with a radius of 1.5
m?
122. An object of mass 3.0 kg is whirled around in a vertical circle of radius 1.3 m with a constant velocity of 6.0
m/s. Calculate the maximum and minimum tension in the string.
123. Snoopy is flying his vintage war plane in a “loop the loop” path chasing the Red Baron. His instruments tell
him the plane is level (at the bottom of the loop) and travelling with a speed of 180 km/h. He is sitting on a set
of bathroom scales, and notes that they read four times the normal force of gravity on him. What is the radius
of the loop? Answer in metres.
124. A pendulum of mass 1.0 kg is suspended from the roof of a car travelling on a level road. An observer in the
car notices that the pendulum string makes an angle of 10° with the vertical. What is the acceleration of the
car?
125. A plane is flying in a vertical loop of 1500 m radius. At what speed is the plane flying at the top of the loop, if
the vertical force exerted by the air on the plane is zero at this point?
126. A 10 kg box is pulled across a level floor, where the coefficient of kinetic friction is 0.35. What horizontal
force is required for an acceleration of 2.0 m/s2?
Dynamics Extra Study Questions
Answer Section
SHORT ANSWER
1. ANS:
(a) Using Newton’s Second Law,
(b) Since the acceleration is constant, the equations of uniformly accelerated motion may be used to describe
its motion.
(c)
REF: K/U
MSC: SP
2. ANS:
(a)
(b)
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.4, p.136
(c)
REF: K/U
MSC: P
3. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.4, p.138
REF: K/U
OBJ: 2.2
LOC: FMV.01
MSC: P
4. ANS:
(a) the northward force of the football on the kicker’s toe
(b) the forward force of the ground on the jogger’s shoe
(c) the upward force of the desk on the book
(d) the forward force of the exhaust gases on a jet engine
(e) the forward force of the water on the swimmer’s hands
KEY: FOP 4.4, p.138
(a)
(b)
REF: C, K/U
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.6, p.144
MSC: P
5. ANS:
For forces to be “balanced,” they must act on the same object and be equal and opposite. The reaction to any
force does not act on the same object as the force; it acts in the opposite direction on the agent exerting the
original force. Thus, when considering the motion of one object, we look at the force, and when considering
the motion of the agent of this force, we look at the reaction.
REF: C, K/U
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.6, p.145
MSC: P
6. ANS:
The fireman has to lean forward. The hose pushes the water forward (the force) so the water pushes backward
on the hose with an equal force (the reaction). To keep from being pushed over backward by the reaction
force, the fireman leans forward.
REF: C, MC
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.6, p.145
MSC: P
7. ANS:
He saves himself by throwing nuts towards the edge of the roof. To do so, he exerts a force on each nut
towards the edge. The nut exerts an equal force back on the squirrel, away from the edge, causing him to slow
down.
REF: C, K/U
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.6, p.145
MSC: P
8. ANS:
If the engine exerts a constant force down on the exhaust gases, the exhaust gases exert a constant force up on
the engines and, hence, the entire rocket. However, as fuel is burned, the rocket’s mass decreases, and since
, the rocket will have a greater acceleration 50 m above the ground than at ground level.
REF: C, MC
MSC: P
9. ANS:
(a)
OBJ: 2.2
LOC: FMV.03
KEY: FOP 4.6, p.145
(b) Consider the 20 kg crate only. The force accelerating it is the force exerted on it by the 12.0 kg crate,
For the reversed crates,
.
is unchanged but
REF: K/U
OBJ: 2.3
MSC: P
10. ANS:
(a) For the entire train,
LOC: FMV.01
KEY: FOP 4.6, p.147
For the force between the locomotive and the 1st car, the total mass being accelerated by this force is 20 cars
of mass 1.0 ×
kg each.
(b) For the force between the 10th and 11th car,
REF: K/U, MC
OBJ: 2.3
MSC: P
11. ANS:
For the acceleration of the entire group,
LOC: FMV.01
KEY: FOP 4.6, p.147
(b) For the force between the 20.0 kg and the 24.0 kg child, mass being accelerated is 16.0 kg + 24.0 kg.
Similarly, for the force between the 24.0 kg and 16.0 kg child,
Free body diagrams of the 3 children appear as:
REF:
K/U
OBJ: 2.3
LOC: FMV.01
KEY: FOP 4.6, p.147
MSC: P
12. ANS:
REF: K/U
MSC: P
13. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.8, p.151
This forward force is exerted on the sprinter by the ground and is the reaction to the backward force that her
feet exert on Earth.
REF: K/U, C
MSC: P
14. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.8, p.151
(a)
(b)
REF: K/U
MSC: P
15. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.8, p.151
REF: K/U
MSC: P
16. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.8, p.152
(a)
(b)
REF: K/U
MSC: P
17. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.8, p.152
Breaking the applied 150 N force into horizontal and vertical components:
The net horizontal force accelerating the mower is:
REF: K/U
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.8, p.153
MSC: P
18. ANS:
If the angle of the between the rope and horizontal is θ, the net horizontal force on the package is:
But
REF: K/U
MSC: P
19. ANS:
For the cart,
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.8, p.153
For no acceleration of the boy, Fnet = 0.
Since the rope exerts a force of 2.0 × 102 N on the boy, his feet must have an opposing force of 2.0 × 102 N.
REF: K/U
MSC: P
20. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.8, p.153
(a)
(b) The force on the cart when a = 0 is 0.60 N. This represents the constant force of friction on the cart.
REF: K/U, I
MSC: P
21. ANS:
OBJ: 2.2
LOC: FM2.01
KEY: FOP 4.8, p.153
REF: K/U
MSC: SP
22. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.4, p.136
REF: K/U
MSC: P
23. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.4, p.138
If the boat pulls on each skier with 400 N, each skier pulls back with 400 N on the boat, as shown. Also, if the
boat pushes back on the water with , the water pushes forward with
on the boat, as shown. Since the
boat is moving in a straight line, at a steady speed,
Taking the components in the direction of the boat,
REF: K/U
MSC: P
24. ANS:
(a)
OBJ: 2.3
LOC: FMV.01
to the plane, and
The force of gravity has two components:
normal force
equal and opposite to
.
KEY: FOP 4.5, p.143
parallel to the plane. The plane exerts a
accelerated the block down the plane.
For the magnitude of
(b)
REF:
K/U
OBJ: 2.3
LOC: FMV.01
KEY: FOP 4.5, p.143
MSC: P
25. ANS:
REF: K/U
MSC: SP
26. ANS:
(a)
OBJ: 2.2
LOC: FM1.01
KEY: FOP 5.1, p.155
OBJ: 2.2
LOC: FM1.01
KEY: FOP 5.1, p.155
REF: K/U
OBJ: 2.2
LOC: FM1.01
MSC: P
28. ANS:
Since the acceleration is zero, Ff = F = 50 N; also, FN = Fg
KEY: FOP 5.1, p.155
(b)
REF: K/U
MSC: P
27. ANS:
(a)
(b)
REF: K/U
MSC: SP
29. ANS:
OBJ: 2.4
LOC: FM1.01
KEY: FOP 5.4, p.163
Note: This force is divided among the four wheels of the car.
REF: K/U
MSC: SP
30. ANS:
OBJ: 2.4
LOC: FMV.01
KEY: FOP 5.4, p.164
REF: K/U
MSC: P
31. ANS:
OBJ: 2.4
LOC: FMV.01
KEY: FOP 5.4, p.165
REF: K/U
MSC: P
OBJ: 2.4
LOC: FMV.01
KEY: FOP 5.4, p.166
32. ANS:
Since g for the Moon is 1.6 N/kg, on the Moon ag will be 1.6 m/s2.
REF: K/U
MSC: SP
33. ANS:
(a)
OBJ: 2.2
LOC: FMV.01
KEY: FOP 5.6, p.168
(b)
REF: K/U
OBJ: 2.2
LOC: FMV.01
KEY: FOP 5.6, p.169
MSC: P
34. ANS:
(a) The time required for the projected rock to reach the ground is equal to the time it would take it to fall
straight down. The time required to fall vertically is determined as follows.
Assume that up is positive and down is negative.
(b) The motion in the horizontal direction is uniform.
Thus,
REF: K/U
OBJ: 1.4
LOC: FM1.03
MSC: SP
35. ANS:
Assuming that the horizontal motion is uniform:
KEY: FOP 5.8, p.172
For the vertical motion, assuming up is positive:
REF: K/U
MSC: P
36. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.172
REF: K/U, MC
MSC: P
37. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.9, p.179
REF: K/U
MSC: P
38. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.9, p.179
Fc is supplied by the tension in the string.
REF: K/U
MSC: P
39. ANS:
OBJ: 3.2
LOC: FM1.04
KEY: FOP 5.10, p.183
REF: K/U
MSC: P
40. ANS:
OBJ: 3.2
LOC: FM1.04
KEY: FOP 5.10, p.183
(a)
(b)
This horizontal displacement is much higher than the actual range would be in practice, since air resistance
has been ignored in the calculation, even at the high speeds involved.
REF: K/U
MSC: SP
41. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.9, p.178
REF: K/U
MSC: P
42. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.174
REF: K/U, MC
MSC: P
43. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.13, p.202
(a)
(b)
REF: K/U, MC
MSC: P
44. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.13, p.202
REF: K/U, C
MSC: P
45. ANS:
OBJ: 3.2
LOC: FM1.04
KEY: FOP 5.13, p.202
OBJ: 3.2
LOC: FM1.04
KEY: FOP 5.13, p.202
REF: K/U
MSC: P
PROBLEM
46. ANS:
For the first interval of motion, when Fa and Ff are acting:
For the second interval of motion, when only Ff is acting:
REF: K/U, I
MSC: P
47. ANS:
REF: K/U
MSC: SP
OBJ: 2.4
LOC: FMV.01
KEY: FOP 4.4, p.138
OBJ: 2.3
LOC: FMV.01
KEY: FOP 4.5, p.139
48. ANS:
Using the vector diagram shown,
REF: K/U
OBJ: 2.3
MSC: P
49. ANS:
(a) Using the vector diagram shown:
LOC: FMV.01
KEY: FOP 4.5, p.143
Using the cosine law:
Using the sine law:
(b)
REF: K/U
MSC: P
50. ANS:
OBJ: 2.3
LOC: FMV.01
KEY: FOP 4.5, p.143
(a) For the flatcar,
(b) The track exerts an equal force in the opposite direction to the sideways force of the rope (actually 5.1 ×
103 N).
REF: K/U, C
MSC: P
51. ANS:
(a)
OBJ: 2.3
LOC: FMV.01
KEY: FOP 4.5, p.143
(b)
(c)
REF: K/U
MSC: P
52. ANS:
OBJ: 2.3
LOC: FMV.01
KEY: FOP 4.6, p.147
The car stopped 3.0 m short of hitting the child.
REF: K/U, MC
MSC: P
53. ANS:
Without the child,
OBJ: 2.3
LOC: FMV.01
KEY: FOP 4.8, p.152
With the child,
The mass of the child is 55 kg – 11 kg = 44 kg.
REF: K/U
MSC: P
54. ANS:
OBJ: 2.3
LOC: FMV.01
KEY: FOP 4.8, p.152
REF: K/U
MSC: P
55. ANS:
For any interval,
OBJ: 2.2
LOC: FMV.01
KEY: FOP 4.8, p.152
REF: K/U, C
OBJ: 2.3
LOC: FMV.02
KEY: FOP 4.8, p.152
MSC: P
56. ANS:
The net force acting on the sled is the vector sun of the three forces acting:
Using components in the x-y direction:
REF: K/U
MSC: P
57. ANS:
(a)
OBJ: 2.3
Then, if
(b) For the second glider,
LOC: FMV.02
KEY: FOP 4.8, p.153
REF: K/U, MC
OBJ: 2.3
LOC: FMV.03
KEY: FOP 4.8, p.153
MSC: P
58. ANS:
(a) The baby carriage is being pushed along a “rough” sidewalk; therefore, there must be some frictional force
that is opposing its motion, between its wheels and the concrete. However, since the carriage is moving with a
constant velocity (zero acceleration), according to Newton’s Second Law the net force acting on it must be
zero. Thus, the other force acting on the carriage is the force of friction, and its value is 200 N in the direction
opposite to the carriage’s motion.
(b) The acceleration of the carriage is given by
The net horizontal force in the direction of this acceleration is
REF: C, K/U
MSC: SP
59. ANS:
OBJ: 2.3
LOC: FMV.01
KEY: FOP 4.4, p.137
REF: K/U
MSC: P
60. ANS:
(a)
(b)
OBJ: 2.3
LOC: FMV.01
KEY: FOP 4.4, p.138
(c)
(d) If v is uniform,
(e) If falling freely, there is no force of the elevator on the passenger, as seen below.
REF: K/U
MSC: P
61. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 5.2, p.158
(a)
(b)
(c)
(d)
(e)
REF:
K/U
OBJ: 2.2
LOC: FMV.02
KEY: FOP 5.2, p.158
MSC: P
62. ANS:
(a)
vertical components:
(b) Acceleration is down, since value of a is negative.
REF: K/U
MSC: P
63. ANS:
OBJ: 2.2
(a) Using Newton’s Law of Motion,
LOC: FMV.02
KEY: FOP 5.2, p.158
(b)
REF: K/U
OBJ: 2.3
MSC: SP
64. ANS:
For m1:
the net force on m1 is
The acceleration of m1 is
For m2:
the net force on m2 is
LOC: FMV.02
KEY: FOP 5.3, p.159
The acceleration of m2 is
The acceleration of the masses have the same magnitude, but opposite directions. Thus
The acceleration of each mass is the same. Substituting for m1 the acceleration is
REF: K/U
MSC: SP
65. ANS:
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.3, p.160
REF: K/U
MSC: P
66. ANS:
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.3, p.161
REF: K/U
MSC: P
67. ANS:
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.3, p.161
REF: K/U
MSC: P
68. ANS:
OBJ: 2.3
Now
Taking the vertical components,
LOC: FMV.02
KEY: FOP 5.3, p.161
But
, since
is less than
.
Thus
As a result, the force of friction is
The net horizontal force on the box is
REF: K/U
MSC: SP
69. ANS:
(a)
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.4, p.164
(b)
REF: K/U
MSC: SP
70. ANS:
OBJ: 2.3, 2.4
LOC: FMV.02
KEY: FOP 5.4, p.165
REF: K/U
MSC: P
71. ANS:
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.4, p.166
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.4, p.166
Since v is constant,
But
REF:
K/U
MSC: P
72. ANS:
(a)
(b)
Taking down the slope as positive
REF: K/U
MSC: P
73. ANS:
(a)
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.4, p.166
(b)
Note that the acceleration and the final speed of the skier do not depend on the mass of the skier.
REF: K/U
MSC: P
74. ANS:
OBJ: 2.4
LOC: FMV.02
At v = 10.3 m/s, the force of air resistance is 3.0 N [up]
KEY: FOP 5.4, p.166
REF: K/U, I
MSC: SP
75. ANS:
(a)
OBJ: 2.4
(b) At terminal velocity,
(c)
LOC: FMV.02
KEY: FOP 5.7, p.170
REF: K/U, I
MSC: P
76. ANS:
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.7, p.171
(a) Consider first the vertical component of the ball’s motion.
When the ball strikes the ground,
The solutions of the equation are
The time taken to reach the ground is 8.0 s, since the negative solution has no meaning in this problem.
(b)
(c) The vertical component of the velocity is
The horizontal component of the velocity is 8.5 m/s. Therefore the resultant velocity is the vector sum of the
vertical and horizontal components as follows:
The velocity of impact is 64 m/s [82° below the horizontal].
REF: K/U
MSC: SP
77. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.173
REF: K/U
MSC: P
78. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.174
REF: K/U
MSC: P
79. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.174
(a)
(b)
REF: K/U
MSC: P
80. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.174
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.174
(a)
(b)
(c)
(d)
REF:
K/U, MC
MSC: P
81. ANS:
(a)
(b)
(c)
REF: K/U, MC
MSC: P
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.174
82. ANS:
(a)
(Negative answer has no meaning in this question.)
(b) Since the balloon and the object are both moving with horizontal velocity of 1.6 m/s, there will be no
horizontal displacement of the object from the balloon. (Relative to the ground there will be a horizontal
displacement.)
REF: K/U, C
MSC: P
83. ANS:
Solution 1:
Solution 2:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.8, p.175
REF: K/U
MSC: P
84. ANS:
OBJ: 1.4
LOC: FM1.03
At the bottom, where the two forces act in opposite directions,
Taking vertical components,
KEY: FOP 5.9, p.178
This is the maximum value for the tension in the rod, pulling up.
At the top, where the two forces act downward,
Taking vertical components,
This is the minimum value for the tension in the rod, pulling down.
Alternate Solution:
At the bottom, the rod must support the ball’s weight as well as provide the centripetal force necessary to
make it move in a circle.
At the top, the ball’s weight provided part of the centripetal force; the rod provides the rest.
REF: K/U
MSC: SP
85. ANS:
(a)
OBJ: 3.2
LOC: FM1.04
KEY: FOP 5.10, p.182
(b)
REF: K/U, MC
MSC: P
86. ANS:
OBJ: 3.2
LOC: FM1.04
KEY: FOP 5.10, p.183
REF: K/U
MSC: P
87. ANS:
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.13, p.197
Alternate Solution:
Since
REF: K/U
MSC: P
88. ANS:
OBJ: 2.2
LOC: FMV.01
KEY: FOP 5.13, p.197
REF: K/U, MC
MSC: P
89. ANS:
(a) (i) Engine On
OBJ: 2.2
LOC: FMV.02
KEY: FOP 5.13, p.197
(ii) Engine Off
(b)
(c)
When the engine was off:
(d)
REF: K/U
MSC: P
90. ANS:
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.13, p.197
REF: K/U
MSC: P
91. ANS:
(a)
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.13, p.197
Note the free body diagram is for the helicopter and the car combined.
(b)
Note the free body diagram is for the car alone.
REF: K/U
MSC: P
92. ANS:
Mass of Jane
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.13, p.197
LOC: FMV.02
KEY: FOP 5.13, p.198
Mass of Tarzan
Mass of Cheetah
Maximum tension
Maximum acceleration of Jane would be:
REF: K/U
MSC: P
OBJ: 2.3
93. ANS:
(a)
(b)
Horizontal Components:
(c)
(d)
(e)
(f) The upward force of the ground is
(Newton’s Third Law).
(g)
REF: K/U
MSC: P
94. ANS:
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.13, p.198
REF: K/U
MSC: P
95. ANS:
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.13, p.198
(b)
(c) No. The net force depends only on the direction of the acceleration and not on the direction of motion.
REF: K/U, C
MSC: P
96. ANS:
(a)
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.13, p.198
(b)
(c) If the cable breaks, there is no force on the object by the scale.
REF: K/U
MSC: P
97. ANS:
(a)
(i)
(ii)
(iii)
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.13, p.198
(iv)
(b)
(i)
(ii)
(iii)
(iv)
(c)
(i)
1.
2.
3.
(ii) From (1):
Note! Do part (iii) first to get a.
From (3):
(iii) The sum of (1) and (2), and (3) is:
(iv)
REF: K/U
MSC: P
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.13, p.198
98. ANS:
(a)
Note: part (b) is done before part (a)
(b)
(a)
(c) For no motion,
and,
and,
REF: K/U
MSC: P
99. ANS:
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.13, p.199
REF: K/U
MSC: P
100. ANS:
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.13, p.199
REF: K/U
MSC: P
101. ANS:
(a)
OBJ: 2.3
LOC: FMV.02
KEY: FOP 5.13, p.199
At a speed of 40
(b)
Note that this acceleration only occurs for a few seconds, when
REF: K/U
MSC: P
102. ANS:
(a)
(b)
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.13, p.200
REF: K/U
MSC: P
103. ANS:
OBJ: 2.4
LOC: FMV.01
KEY: FOP 5.13, p.200
REF: K/U
MSC: P
104. ANS:
Consider top block.
When
OBJ: 2.4
LOC: FMV.01
KEY: FOP 5.13, p.200
Consider the whole system.
REF: K/U
MSC: P
105. ANS:
OBJ: 2.4
(a) Taking vertical components:
(since
LOC: FMV.02
)
KEY: FOP 5.13, p.200
(b)
(c) Taking horizontal components:
REF: K/U
MSC: P
106. ANS:
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.13, p.200
REF: K/U
MSC: P
107. ANS:
OBJ: 2.4
LOC: FMV.01
KEY: FOP 5.13, p.200
Taking components perpendicular to the slope (up positive)
Taking components parallel to the slope (up positive)
REF: K/U
MSC: P
108. ANS:
(a)
OBJ: 2.4
LOC: FMV.02
KEY: FOP 5.13, p.200
(b)
(c) (i)
(ii)
(d) Air resistance increases with velocity.
Since
increases,
decreases and a decreases.
REF: K/U, C
MSC: P
109. ANS:
OBJ: 2.4
LOC: FM2.01
KEY: FOP 5.13, p.200
REF: K/U
MSC: P
110. ANS:
(a)
OBJ: 2.3
LOC: FMV.01
KEY: FOP 5.13, p.201
(b)
Note that this force is about seven times his weight—half from each leg.
REF: K/U
MSC: P
111. ANS:
OBJ: 2.3
LOC: FMV.01
KEY: FOP 5.13, p.201
REF: K/U
MSC: P
112. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.13, p.201
REF: K/U, MC
MSC: P
113. ANS:
(a)
(b)
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.13, p.201
(c)
REF: K/U
MSC: P
114. ANS:
(a)
(b)
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.13, p.201
(c)
REF: K/U
MSC: P
115. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.13, p.201
(a)
(b) no air resistance
no friction car and bridge
no reduction in speed because of guardrail, etc.
vertical motion independent of horizontal motion
(c) Underestimate, since car would have travelled farther horizontally without the above.
REF: K/U, MC, I
MSC: P
116. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.13, p.201
(a)
or
(b)
REF: K/U
MSC: P
117. ANS:
Total time:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.13, p.202
REF: K/U
MSC: P
118. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.13, p.202
Vertical time of flight:
Distance for second player to run is 30 m – 22.8 m = 7.2 m.
REF: K/U, MC
MSC: P
119. ANS:
OBJ: 1.4
LOC: FM1.03
KEY: FOP 5.13, p.202
OBJ: 3.2
LOC: FM1.04
KEY: FOP 5.13, p.202
(a)
(b)
REF: K/U
MSC: P
120. ANS:
(a)
and
(b)
REF: K/U
MSC: P
121. ANS:
OBJ: 3.2
LOC: FM1.04
KEY: FOP 5.13, p.203
At the top,
At the top, T = 0
REF: K/U
OBJ: 3.2
MSC: P
122. ANS:
Note the mass of the ball is immaterial.
At the top,
At the bottom,
LOC: FM1.04
KEY: FOP 5.13, p.203
REF: K/U
MSC: P
123. ANS:
OBJ: 3.2
REF: K/U
OBJ: 3.2
MSC: P
124. ANS:
Ball and car accelerating at same rate.
LOC: FM1.04
KEY: FOP 5.13, p.203
LOC: FM1.04
KEY: FOP 5.13, p.203
Taking vertical components,
Taking horizontal components,
Combining (1) and (2)
REF: K/U, MC
MSC: P
125. ANS:
OBJ: 3.2
LOC: FM1.04
No force is exerted by the air on the plane at the top of the loop.
KEY: FOP 5.13, p.203
REF: K/U
MSC: P
126. ANS:
OBJ: 3.2
LOC: FM1.04
KEY: FOP 5.10, p.183
REF: K/U
MSC: P
OBJ: 2.4
LOC: FMV.01
KEY: FOP 5.13, p.200