Ch 4 Forces & the Laws of Motion
Updated 08/22/07
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Practice A, Drawing Free-Body Diagrams, pg 124
1] Each diagram should include all forces acting on the object, pointing in the
correct direction and with the lengths roughly proportional to the magnitudes of
the forces. Be sure each vector is labeled.
2] Diagrams should include downward gravitational force and upward force of the
desk on the book; both vectors should have the same length and should be
labeled.
Section 4.1 Review Q, pg 124:
1] Answers will vary but should not simply repeat Fig 1.
a] Fig 1: a person throws a ball.
b] Fig 1: a person catches a ball.
c] Fig 1: a bat hits a ball.
2] Gravity and electric force are field forces because they can cause a change
in motion. Forces observed in everyday life: answers will vary. Ex: car
engine and breaks, person jumping or walking
3] Force = mass x acceleration
1 Newton = 1 kilogram x 1 meter per second squared.
1 N = 1 kg x 1 m/s2
Newton = the force needed to accelerate 1 kg at 1 m/s2
4] because force has both magnitude and direction
5] Fg points down, and Fkicker points in the direction of the kick.
6] Each arrow should have a label identifying the object exerting the force and
the object acted on by the force.
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Practice B, Determining Net Force, pg 128
1] Fx = 60.65 N, Fy = 35.0 N
2] 2.48 N at 25o counterclockwise from straight down
3] 357 N at 35.7o west of north
Section 4.2 Review Q, pg 129:
1] zero
2] -3674 N
3] 4502 N at 1.655o forward of the side.
4] The same magnitude as the net force in item 3 but in the opposite direction
5] No, either no force or two or more forces are required for equilibrium.
Ch 4 Review Q, Forces and Newton’s First Law, pg 145:
1] Yes; the object could move at a constant velocity.
2] No, just that the net force equals zero.
3] No; it has a net force downward (gravitational force).
4] mass
5] The ball moves toward the back of the truck because inertia keeps it in place
relative to the ground.
6a] the inertia of the crate
b] It could continue forward by inertial and hit the cab.
7] Fg (8.9 N0 and Fapplied (2.1 N) pointdownward, and Fn (11.0 N) points upward.
8] Fapplied (185 N) points forward, Fg (155 N) points downward, and Fn (155 N)
points upward. The diagram should also include Ffriction backward.
9a] Fg points down, and Fr points up
b] Frotos points up and Fg points down
c] Fg points down, Ftrack points in the direction of motion, and Fn points up.
10] 1210 N at 62o above the 1520 N force
11a] F1 (220 N) and Fx (114 N) point right
F1 (220 N) points left, and F2 (114 N) points right.
b] First situation: 220 N to the right, 114 N to the right
Second situation: 220 N to the left, 114 N to the right
12] 4 N; 3 N
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Practice C, Newton’s 2nd Law, pg 132
[1] 2.2 m/s2 forward [2] 1.4 m/s2 north [3] 4.50
m/s2 East
[4] 2.1 kg [5] 14 N
Section 4.3 Review Q, pg 134:
1a] 12 N [b] 3.0 m/s2
2] The reaction force acts on the child, not on the wagon itself, so there is
still a net force on the wagon.
3a] person pushes on ground; ground pushes on person
b] snowball exerts force on back; back exerts force on snowball
c] ball exerts force on glove; glove exerts force on ball
d] wind exerts force on window; window exerts force on wind.
4] 1.6 m/s2 at 65o north of east
5] Each impact force has the same magnitude. The sports car experiences the
larger acceleration because it has a smaller mass, and acceleration is inversely
proportional to mass.
Ch 4 Review Q, Newton’s 2nd & 3rd Laws, pg 146:
13] because Earth has a very large mass
14] An object with greater mass requires a larger force for a given acceleratin.
15] One-sixth of the force needed to lift an object on Earth is needed on the
moon.
16] On the horse: the force of the cart, Fg down, Fn up, a reaction force of
the ground on the hooves.
On the cart: the force of the horse, Fg down, Fn up, kinetic friction
17] Push it gently; with a smaller force, the astronaut will experience a
smaller reaction force.
18] As the climber exerts a force downward, the rope supplies a reaction force
that is directed upward. When this reaction force is greater than the climber’s
weight, the climber accelerates upward.
19a] zero [b] zero
20] 3.52 m/s2
21] 55 N to the right
22a] 770 N at 8.1o right of forward
b] 0.24 m/s2 at 8.1o right of forward
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Practice D, Coefficients of Friction, pg 139
[1] 0.23 [2a] 0.67 [b] 0.52
[3a] 870 N, 670 N [b] 110 N, 84 N [c] 1000 N, 500 N [d] 5 N, 2 N
Practice E, Overcoming Friction, pg 141
1] 2.7 m/s2 in the positive x direction
2] 0.77 m/s2 up the ramp
3a] 0.061 b] 3.61 m/s2 down the ramp
Note: m cancels in the solution, and a is the same in both cases; the
slight difference is due to rounding.
4] 0.609
Section 4.4 Review Q, pg 143:
1a] An arrow labeled Fg should point down, and an arrow labeled Fair should
point opposite the direction of motion. The Fg arrow should be longer then the
Fair arrow.
b] Fg points down, Fn points up, Fapplied is horizontal, and Ffriction points
in the opposite direction. The two vertical arrows are equal in length, as are
the two horizontal arrows.
2a] 3.70 N [b] 58.5 N
3a] 34 N [b] 39 N
4] 0.37, 0.32
5] No; once at equilibrium the velocity will not increase, so the force of air
resistance will not increase.
Ch 4 Review Q, Weight, Friction, & Normal Force, pg
146:
23] Mass is the inertial property of matter. Weight is the gravitational force
acting on an object. Weight is equal to mass times the free-fall acceleration.
24a] -1.47 N [b] -1.47 N]
25a] Fg points down, and Fn points up
b] Fg points down, and Fn points up perpendicular to ramp’s surface
c] same as [b] [d] same as [b]
26a] 54 N [b] 53 N [c] 49 N [d] 38 N
27] The normal force decreases. The force of static friction increases to
counteract the component of the weight along the table.
28] Fs max [29] 51 N
30] 0 N; the forces exerted by each star cancel.
31a] The weight of the ball and an equal reaction force of the ball
on Earth. The force of the person’s hand on the ball and an
equal reaction force of the ball on the hand.
b] Fg; the force of the ball on Earth
32] Pushing down on the book increases the normal force and therefore also
increases the friction.
33] The rock will accelerate until the size of the resistive force equals the
net downward force on the rock. (This downward force is the rock’s weight minus
the buoyant force of the water.) Tehn the rock’s speed will be constant.
34] As the sky diver’s speed increases, the acceleration decreases because the
resistive force increases with increasing speed; zero
35] 0.70, 0.60 [36] 0.436 [37] 0.816
38] 1.4 m/s2 down the aisle [39] 1 m/s2
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40] 68 N; 34 N
41] 13 N down the incline
42a] 9.81 m/s2 downward [b] 22.2 N
43] 64 N up
44a] zero [b] 33.9 N
45a] 0.25 m/s2 forward [b] 18 m [c] 3.0 m/s
46] 50 m
47a] 2 s [b] The box will never move. The force exerted is not enough to
overcome friction.
48a] 1.78 m/s2 [b] 0.37 [c] 9.4 N [d] 2.67 m/s
49] -1.2 m/s2; 0.12
50] -500 N
51a] 2690 forward [b] 699 N forward
52] 32.2 N
53] 13 N, 13 N, 0 N, -26 N
54] 1.41o
1] C [2] G [3] C [4] G [5] A [6] G [7] A [8] F
[9] D
10] 6.00 s
11] 72.0 m
12] 63.6 m/s
13] At rest moves left, hits back wall
14] Moves right (with velocity v), at rest, neither
15] Moves right, moves right, hits front wall
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