Exercise 22 Solution
1. Which of
Newton's laws best explains why motorists should buckleup?
A) the first law
B) the second law
C) the third law
D) the law of gravitation
Simole illustration of the law
2. A rocket moves through empty
space in a straight line with constant speed.
It is far from the
gravitational effect of any star or planet.
Under these conditions, the force that must be applied to
the rocket in order to sustain its motion is
A) equal to its weight.
B) equal to its mass.
C) dependent on how fast it is moving.
D) zero.
Newton's first law
3. You are standing in a moving
bus, facing forward, and you suddenly fall forward as the bus comes
to an immediate stop. What
force caused you to fall forward?
A) gravity
B) normal force due to your contact with the floor of the bus
C) force due to friction between you and the floor of the bus
D) There is not a force leading to your fall.
It is no longer an inertial reference frame (one that is in motion with constant velocity or one that is not in motion). Newton's laws hold only
in inertial reference frames.
4. A constant net force acts on
an object. Describe the motion
of the object.
A) constant acceleration
B) constant speed
C) constant velocity
D) increasing acceleration
in this case
5. A net force F accelerates a
mass m with an acceleration a.
If the same net force is applied to
mass 2m, then the acceleration will be
A) 4a.
B) 2a.
C) a/2.
D) a/4.
If m is doubled, then a is divided by 2
6. If you blow up a balloon,
and then release it, the balloon will fly away.
This is an illustration of
A) Newton's first law.
B) Newton's second law.
C) Newton's third law.
D) Galileo's law of inertia.
Simple illustration of the law
7. A 20ton truck collides with
a 1500lb car and causes a lot of damage to the car.
Since a lot of
damage is done on the car
A) the force on the truck is greater than the force on the car.
B) the force on the truck is equal to the force on the car.
C) the force on the truck is smaller than the force on the car.
D) the truck did not slow down during the collision.
The force is the same by Newton's third law. Since F = ma, the one with the smaller mass attains a higher acceleration with the impact, resulting in more damage.
8. A golf club hits a golf ball
with a force of 2400 N. The
golf ball hits the club with a force
A) slightly less than 2400 N.
B) exactly 2400 N.
C) slightly more than 2400 N.
D) close to 0 N.
Same as problem 8
9. Mass and weight
A) both measure the same thing.
B) are exactly equal.
C) are two different quantities.
D) are both measured in kilograms.
10. A stone is thrown straight up.
At the top of its path, the net force acting on it is
A) greater than its weight.
B) greater than zero, but less than its weight.
C) instantaneously equal to zero.
D) equal to its weight.
11. A 20N weight and a 5.0N weight are dropped simultaneously from the
same height. Ignore air
resistance. Compare their
accelerations.
A) The 20 N weight accelerates faster because it is heavier.
B) The 20 N weight accelerates faster because it has more inertia.
C) The 5.0 N weight accelerates faster because it has a smaller mass.
D) They both accelerate at the same rate because they have the same weight
to mass ratio.
F = ma 20 = m_{1}a 5 = m_{2}a since the accelerations are the same, 20/m_{1} = 5/m_{2}
Saying they have the same weight to mass ratio is synonomous with saying they have the same acceleration  which is true
12. An object of mass m is hanging by a string from the ceiling of an
elevator. The elevator is
moving
up at constant speed. What is
the tension in the string?
A) less than mg
B) exactly mg
C) greater than mg
D) cannot be determined without knowing the speed
13. A block of mass M slides down a frictionless plane inclined at an angle
θ with the horizontal. The
normal reaction force exerted by the plane on the block is
A) Mg.
B) Mg sin θ.
C) Mg cos θ.
D) zero, since the plane is frictionless.
Draw a freebody diagram to see this
14. A block of mass M slides down a frictionless plane inclined at an angle
θ with the horizontal. The
gravitational force is directed
A) parallel to the plane in the same direction as the movement of the block.
B) parallel to the plane in the opposite direction as the movement of the
block
C) perpendicular to the plane.
D) toward the center of the Earth.
15. Two toy cars (16 kg and 2.0 kg) are released simultaneously on an
inclined plane that makes an
angle of 30° with the horizontal.
Make a statement which best describes their acceleration after
being released.
A) The 16kg car accelerates 8 times faster than the 2.0kg car.
B) The 2.0kg car accelerates 8 times faster than the 16kg car.
C) Both cars accelerate at the same rate.
D) none of the above
Galileo showed this with his experiments  which we duplicated in a lab
16. An object sits on a frictionless surface.
A 16N force is applied to the object, and it accelerates at
2.0 m/s2.
What is the mass of the object?
A) 4.0 kg
B) 8.0 kg
C) 32 kg
D) 78 N
17. Starting from rest, a 4.0kg body reaches a speed of 8.0 m/s in 2.0 s.
What is the net force acting
on the body?
A) 4.0 N
B) 8.0 N
C) 16 N
a = change in speed/change in time = (8  0)/(2  0) = 4
F = ma = 4x4 = 16
18. Sue and Sean are having a tugofwar by pulling on opposite ends of a
5.0kg rope. Sue pulls
with a 15N force. What is
Sean's force if the rope accelerates toward Sue at 2.0 m/s2?
A) 3.0 N
B) 5.0 N
C) 25 N
D) 50 N
It accelerates towards Sue, so her force must exceed that of Sean's
F_{Sue}  F_{Sean} = m_{rope}a_{rope }
19. A traffic light of weight 100 N is supported by two ropes as shown in the
figure
on the right.
A) 50 N
B) 63 N
C) 66 N
D) 83 N


Sum forces in y direction and obtain F_{Ty} = 50 N F_{T} = F_{Ty}/sin37 = 50/0.602 = 83 N
20. Two boxes of masses m and 2m are in contact with each other on a frictionless surface. (See figure on the right). What is the acceleration of the
more
massive box?
A) F/m
B) F/(2m)
C) F/(3m)
D) F/(4m)
F = ma gives a = F/


21. A 10kg box sitting on a horizontal surface is pulled by a 5.0N force.
A 3.0N friction force retards
the motion. What is the
acceleration of the object?
A) 0.20 m/s^{2}
B) 0.30 m/s^{2}
C) 0.50 m/s^{2}
D) 5.0 m/s^{2}
22. A horizontal force of 5.0 N accelerates a 4.0kg mass, from rest, at a
rate of 0.50 m/s^{2} in
the
positive direction. What
friction force acts on the mass?
A) 2.0 N
B) 3.0 N
C) 4.0 N
D) 5.0 N
23. A 10kg mass slides down a flat hill that makes an angle of 10° with the
horizontal. If friction is
negligible, what is the resultant force on the sled?
A) 1.7 N
B) 17 N
C) 97 N
D) 98 N
Draw a free body diagram and find the component of the gravitational force in the x direction
24. A mass is placed on a smooth inclined plane with an angle of 37° to the
horizontal. If the inclined
plane is 5.0m long, how long does it take for the mass to reach the bottom
of the inclined plane
after it is released from rest?
A) 1.3 s
B) 1.2 s
C) 1.1 s
D) 1.0 s
Find F in x direction using angles, then set F = ma, find that a_{box}
= 5.9 m/s^{2}
use x = 1/2 at^{2 }and solve for t
25. An object with a mass m slides down a rough 37° inclined plane where the
coefficient of kinetic
friction is 0.20. What is the
acceleration of the object?
A) 4.3 m/s^{2}
B) 5.9 m/s^{2}
C) 6.6 m/s^{2}^{ }
D) 7.8 m/s^{2}^{ }
Find F of gravity in rotated x direction and subtract from it F_{fr}.
Set equal to ma_{box}
26. An object with a mass m slides down a rough 37° inclined plane where the
coefficient of kinetic
friction is 0.20. If the plane
is 10 m long and the mass starts from rest, what will be its speed at
the bottom of the plane?
A) 12 m/s
B) 11 m/s
C) 9.7 m/s
D) 9.3 m/s
Normal force = mgcos 37
F_{Fr }= mgcos 37x0.20
Sum forces in x direction and set equal to ma_{box}
27. A bulldozer drags a log weighing 500 N along a rough surface.
The cable attached to the log
makes an angle of 30.0° with the ground. The
coefficient of static friction between the log and the
ground is 0.500. What minimum
tension is required in the cable in order for the log to begin to
move?
A) 224 N
B) 268 N
C) 289 N
D) 500 N
Sum forces in y direction (positive is up): Fsin30 + F_{N}  500 = 0 (because no acceleration in y direction)
F_{N} = 500  Fsin30 = 500  0.5F
F_{N} = 500  0.5F
F_{Fr} = 0.5(500  0.5F) = 250  0.25F
To just move F_{Fr} < F
F_{Fr} = F
gives 250  0.25F = Fcos 30
250  0.25 F = 0.866 F
1.116F = 250 gives F = 224
So, anything over 224