MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 3: DYNAMICS
WORKSHEET SOLUTIONS
1. There are four books in a stack, each with a weight of 4 N. The coefficient of friction between each
pair of books is 0.2. With what horizontal force must one push to start sliding the top books off the
bottom one?
1A.
Fx: F – f

Fy: n – w)
(1) Fx: FNET = ma = m(0)  F – f = 0 (there is no acceleration here, that is what “start sliding” means)
(2) F = f
(3) Fy: FNET = ma = m(0)  n – w = 0
(4) n = w
Solve for friction:
(5) f = n
(6) f = (0.2)(12 N)
(7) f = 2.4 N
(8) F = 2.4 N
2. As a 5.0 kg bucket is being lowered into a 20 m deep well, starting from the top, the tension in the
rope is 17 N. What is the time to reach the bottom?
2A.
Fx: 0
(1)
(2)
(3)
(4)

Fy: T – w)
Fx: 0
Fy: FNET = ma = m(- a)  T – w = - ma
- ma = T – w
ma = - (T – w)
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
2A. (continued)
(5) ma = w – T
(6) a = (w – T) / m
Determine the weight of the bucket:
(7) w = mg
(8) w = (5.0 kg)(9.8 m/s2)
(9) w = 49 N
Calculate the acceleration:
(10) a = (49 N – 17 N) / 5.0 kg
(11) a = 6.4 m/s2
Determine the time:
(12) d = vot + ½ at2
(13) d = 0 + ½ at2
(14) ½ at2 = d
(15) t = (2d/a)1/2
(16) t = [2(20 m) / 6.4 m/s2]1/2
(17) t = 2.5 s
3. A woman at an airport is pulling a 7 kg suitcase with wheels at constant speed by pulling on a strap at
an angle  above the horizontal. She pulls on the strap with a 30 N force, and the frictional force is 25
N. What is the angle ?
3A.
(Fx: Tcos – f
Fy: n – w + Tsin)
(1) Fx: FNET = ma = m(0) T cos  – f = 0 (when a problem says “constant speed”, that means no
acceleration)
(2) T cos f
(3) Fy: FNET = ma = m(0)  T sin  + n – w = 0
(4) T sin n = w
(5) T cos  = f (from Fx)
(6) cos  = f / T
(7) cos  = 25 N / 30 N
(8) cos  = 0.833
(9)  = arc cos (0.833)
(10)  = 33.6o
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
4. Two unequal masses are falling through the air with the heavier one below the lighter one. They are
connected by a string. (Ignore air friction). The tension in the string will be equal to:
(A) the weight of the heavier mass
(B) the weight of the lighter mass
(C) the difference in weight of the two masses
(D) the mass of both objects
(E) zero
4A.
(E) zero. Both objects are in free-fall motion. Compared to the massiveness of the earth, their masses
don’t matter in this calculation. They both fall at exactly 9.8 m/s2.
5. A 12 kg block rests on a level frictionless surface and is attached by a light string to a 7 kg hanging
mass where the string passes over a massless frictionless pulley. The tension in the string is 60 N. If g
= 9.8 m/s2, what is the acceleration of the system when released?
5A.
(1) FOR MASS 1:
Fx: FNET = ma  T = m1a
Fy: FNET = ma = m(0)  n – w1 = 0
(2)
FOR MASS 2:
Fx: 0
Fy: FNET = ma = m(- a) T – w2 = - m2a
(3) T = m1a
(4) a = (T / m1)
(5) a = (60 N / 12 kg)
(6) a = 5 m/s2
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
6. A puck, upon being struck by a hockey stick, is given an initial speed of 8 m/s and continues to move
in a straight path for a distance of 16 m before coming to rest. What is the coefficient of kinetic friction
between puck and ice?
6A.
Hockey Puck (v = 0)
(Fx: F – f
Fy: n - w)
(1) Fx: FNET = ma = m(0)  F – f = 0
(2) F = f
(3) Fy: FNET = ma = m(0)  n – w = 0
(4) n = w
Choose a kinematics equation:
(5) v2 = vo2 + 2ad
(6) 0 = vo2 + 2ad
(7) - vo2 = 2 ad
(8) a = - (vo2) / 2d
(9) a = - (8 m/s)2 / 2 (16 m)
(10) a = - 2 m/s2
(11) F = f (from Fx)
(12) F = ma
(13) f = n
(14) ma = n
(15) ma = w
(16) ma = (mg)
(17) a = g
(18)  = a / g
(19)  = (- 2 m/s2) / (- 9.8 m/s2)
(20)  = 0.204
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
7. Find the tension in an elevator cable if the 1000 kg elevator is descending with an acceleration of
1.6m/s2, downward.
7A.
(Fx: 0
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
Fy: T – w)
Fx: 0
Fy: FNET = ma = m(- a)  T - w = - ma
T = w – ma
Weight equation: w = mg
T = (mg) – ma
T = m (g – a)
T = (1000 kg)(9.8 m/s2 – 1.6 m/s2)
T = (1000 kg)(8.2 m/s2)
T = 8,200 N
8. Two perpendicular forces, one of 30 N directed due north; and the second, 40 N directed due east, act
simultaneously on an object with mass of 35 kg. What is the magnitude of the resultant acceleration of
the object?
8A.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
FNET = [(Fx)2 + (Fy)2]1/2
FNET = [(40 N)2 + (30 N)2]1/2
FNET = 50 N
FNET = ma
a = FNET / m
a = 50 N / 35 kg
a = 1.43 m/s2
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
9. Two blocks of masses 4 and 6 kg, respectively, rest on a frictionless horizontal surface and are
connected by a string. A second string attached only to the 6 kg block, has a horizontal force of 20 N
applied to it, causing both blocks to accelerate. What is the tension in the string between the two
blocks?
9A.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
(9)
FNET = ma
FNET = (m1 + m2)a
a = FNET / (m1 + m2)
a = 20 N / (4 kg + 6 kg)
a = 2 m/s2
FNET = ma
F1 = m1a
F1 = (4 kg)(2 m/s2)
F1 = T1 = 8 N
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 3: DYNAMICS
WORKSHEET SOLUTIONS 2
1. A 10 N text book is at rest on the horizontal surface of a lecture table.
1a. Draw a free body diagram for the text book.
A.
(Fx: 0
Fy: n – w)
1b. How are these two forces related?
A.
Since the book is at rest, the net force must be zero. In other words, F = 0. n = w
1c. What is the force the book exerts on the table?
A.
By Newton’s third law, this force must be equal and opposite to the force the table exerts on the book:
10 N downward.
1d. Suppose the lecture table, with the book on it, is accelerated upward at a rate of 1.0 m/s2 (remember
that g = 9.8 m/s2). What is the net force on the book now?
A.
(1) w = mg
(2) m = w /g
(3) m = 10 N / 9.8 m/s2
(4) m = 1.02 kg
(5) FNET = ma
(6) FNET = (1.02 kg)(1.0 m/s2)
(7) FNET = 1.02 N
1e. While accelerating upward, what is the force the table exerts on the book?
A. 11.02 N. The upward force plus the 10 N normal force must equal 1.02 N, the net force on the book.
1f. While accelerating upward, what is the force the book exerts on the table?
A. 11.02 N downward.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
2. Calculations of weight from mass and mass from weight.
2a.
A.
(1)
(2)
(3)
What is the weight of a 1 kg mass on earth, in Newtons (N)?
w = mg
w = (1 kg)(9.8 m/s2)
w = 9.8 N
2b. Given the conversion factor, 1 lb = 4.45 N, calculate the weight of a 1 kg mass in lb.
A.
9.8 N
1 lb
-------- x --------- = 2.2 lb
1
4.45 N
2c. A car weighs 3200 lb. What is its mass in kg?
A. 3200 lb
1 kg
--------- x ------- = 1455 kg
1
2.2 lb
2d. If the car above were taken to the moon’s surface, where the acceleration of gravity is 1/6 its value
on earth, what would be its mass in kg and its weight in lbs?
A.
Its mass is 1455 kg. The car’s weight is 533 pounds (one sixth its value on earth).
3. A box of mass 10 kg is pushed across a horizontal floor by a horizontal force of 60 N. Friction
between the floor and box is 40 N.
Question 3. Diagram
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
3a. Draw a free body diagram.
A.
(Fx: F – f
3b.
A.
(1)
(2)
(3)
3c.
A.
(1)
(2)
(3)
(4)
Fy: n – w)
What is the net force?
Fx: FNET = F – f
FNET = 60 N – 40 N
FNET = 20 N to the right
What is the acceleration of the box?
FNET = ma
a = FNET / m
a = 20 N / 10 kg
a = 2 m/s2
4. A crate of mass 10 kg is pulled up a 37 degree incline by a rope under tension T, directed parallel to
the incline. There is no friction between the crate and the incline (rollers are used).
Question 4. Diagram
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
4a. Draw a free body diagram.
A.
( Fx: T – w(sin 37o) Fy: n – w(cos 37o) )
4b. What value of T is required to hold the crate at rest on the incline?
A.
(1) Fx: FNET = ma = m(0)  T – w(sin 37o) = 0
(2) T = w (sin 37o)
(This is the component of the crate’s weight sliding down the incline)
(3) T = mg (sin 37o)
(4) T = (10 kg)(9.8 m/s2)(0.6018)
(5) T = 59 N
4c. If T is such as to cause the crate to accelerate up the incline at 0.50 m/s2, what is the net force on the
crate?
A.
(1) FNET = ma
(2) FNET = (10 kg)(0.50 m/s2)
(3) FNET = 5 N up the incline
4d. What is the tension in the situation of question 4c?
A. 59 N + 5 N = 64 N.
4e. If T is such as to cause the crate to accelerate down the incline at 0.50 m/s2, what is the net force on
the crate?
A.
(1) FNET = ma
(2) FNET = (10 kg)(0.50 m/s2)
(3) FNET = 5 N down the incline
4f. What is the tension in the situation of question 4e?
A. 59 N – 5 N = 54 N.
(This must be less than 59 N, the component of the weight directed down the incline.)
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
5. Two blocks are connected by a string that runs over a frictionless pulley. There is no friction
between the block and the table’s horizontal surface. T is the tension in the string.
Question 5. Diagram
5a. Draw a free body diagram for mass #1.
A.
(Fx: T
5b. Draw a free body diagram for mass #2.
A.
Fy: n – w1)
(Fx: 0
Fy: T – w2)
5c.
A.
(1)
(2)
(3)
What is the net force on mass #1? Note that the acceleration is given in the figure.
FNET = ma
FNET = (3.0 kg)(1 m/s2)
FNET = 3 N
5d. What is the tension T?
A. F = T = 3 N
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
5e. What is mass #2, in kg?
A.
Fx: 0

Fy: T – w2)

Fy: FNET = ma = m(- a)  T – w2 = - m2a
(2) T = w2 – m2a
(3) T = m2g – m2a
(4) T = m2 (g – a)
(5) m2 = T / (g – a)
(6) m2 = (3.0 N) / (9.8 m/s2 – 1.0 m/s2)
(7) m2 = 0.34 kg
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 3: DYNAMICS
QUIZ SOLUTIONS
1. A woman at an airport is pulling a 10 kg suitcase with wheels at constant speed by pulling on a strap
at an angle  above the horizontal. She pulls on the strap with a 40 N force, and the frictional force is 22
N. What is the angle ?
1A.
(Fx: Tcos – f
Fy: n – w + Tsin)
(1) Fx: FNET = ma = m(0) T cos  – f = 0 (when a problem says “constant speed”, that means no
acceleration)
(2) T cos f
(3) Fy: FNET = ma = m(0)  T sin  + n – w = 0
(4) T sin n = w
(5) T cos  = f (from Fx)
(6) cos  = f / T
(7) cos  = 22 N / 40 N
(8) cos  = 0.55
(9)  = 56.6o
2. A 12 kg block rests on a level frictionless surface and is attached by a light string to a 4 kg hanging
mass where the string passes over a massless frictionless pulley. The tension in the string is 49 N. If g
= 9.8 m/s2, what is the acceleration of the system when released?
2A.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
(1) FOR MASS 1:
Fx: FNET = ma  T = m1a
Fy: FNET = ma = m(0)  n – w1 = 0
(2) FOR MASS 2:
Fx: 0
Fy: FNET = ma = m(- a) T – w2 = - m2a
(3) T = m1a
(4) a = (T / m1)
(5) a = (49 N / 12 kg)
(6) a = 4.08 m/s2
3. A puck, upon begin struck by a hockey stick, is given an initial speed of 9 m/s and continues to move
in a straight path for a distance of 17 m before coming to rest. What is the coefficient of kinetic friction
between puck and ice?
3A.
Hockey Puck (v = 0)
(Fx: F – f
Fy: n - w)
(1) Fx: FNET = ma = m(0)  F – f = 0
(2) F = f
(3) Fy: FNET = ma = m(0)  n – w = 0
(4) n = w
Choose a kinematics equation:
(5) v2 = vo2 + 2ad
(6) 0 = vo2 + 2ad
(7) - vo2 = 2 ad
(8) a = - (vo2) / 2d
(9) a = - (9 m/s)2 / 2 (17 m)
(10) a = - 2.38 m/s2
(11) F = ma
(12) f = n
(13) ma = n (because F = f)
(14) ma = w (because n = w)
(15) ma = (mg) (because w = mg)
(16) a = g
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
3A. (continued…)
(17)  = a / g
(18)  = (- 2.38 m/s2) / (- 9.8 m/s2)
(19)  = 0.243
4. Find the tension in an elevator cable if a 850 kg elevator is descending with an acceleration of 1.6
m/s2, downward.
4A.
(Fx: 0
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
Fy: T – w)
Fx: 0
Fy: FNET = ma = m(- a)  T - w = - ma
T = w – ma
Weight equation: w = mg
T = (mg) – ma
T = m (g – a)
T = (850 kg)(8.2 m/s2)
T = 6970 N
5. There are five books in a stack, each with a weight of 9 N. The coefficient of friction between each
pair of books is 0.2. With what horizontal force must one push to start sliding the top books off the
bottom one?
5A.
Fx: F – f

Fy: n – w)
(1) Fx: FNET = ma = m(0)  F – f = 0 (there is no acceleration here, that is what “start sliding” means)
(2) F = f
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
5A. (continued)
(3) Fy: FNET = ma = m(0)  n – w = 0
(4) n = w
Solve for friction:
(5) f = n
(6) f = (0.2)(36 N)
(7) f = 7.2 N
(8) F = 7.2 N (since F = f)
6. Two blocks of masses 3 and 5 kg, respectively, rest on a frictionless horizontal surface and are
connected by a string. A second string attached only to the 5 kg block, has a horizontal force of 20 N
applied to it, causing both blocks to accelerate. What is the tension in the string between the two
blocks?
6A.
(1)
(2)
(3)
(4)
(5)
(6)
(7)
(8)
FNET = ma
FNET = (m1 + m2)a
a = FNET / (m1 + m2)
a = 20 N / (3 kg + 5 kg)
a = 2.5 m/s2
F1 = m1a
F1 = (3 kg)(2.5 m/s2)
F1 = T1 = 7.5 N
Questions 7 - 10. Calculations of weight from mass and mass from weight:
7. What is the weight of a 3.5 kg mass on earth, in Newtons [N]? Recall that g = 9.8 m/s2.
7A.
(1) w = mg
(2) w = (3.5 kg)(9.8 m/s2)
(3) w = 34.3 N
8. Calculate the weight of a 4 kg mass in lb.
8A.
(1) w = mg
(2) w = (4 kg)(9.8 m/s2)
(3) w = 39.2 N
(4) 39.2 N
1 lb
--------- x --------- = 8.81 lb
1
4.45 N
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
9. A car weighs 2,900 pounds on earth. What is its mass in kg?
9A.
2,900 lb
1 kg
---------- x ------- = 1319 kg
1
2.2 lb
10. If the car above were taken to the moon’s surface, where the acceleration of gravity is 1/6 its value
on earth, what would be its weight in pounds?
10A.
The car’s weight is 483.3 pounds.
Questions 11 - 15. Two blocks are connected by a string that runs over a frictionless pulley (the string
slides without friction so the pulley does not rotate).
Questions 11 – 15. (continued…)
There is no friction between the block and the table’s horizontal surface. T is the tension in the string.
m1 (on table) = 2.5 kg, a = 0.8 m/s2, no friction).
11. Draw a free body diagram for mass #1.
11A.
12. Draw a free body diagram for mass #2.
A.
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MR. SURRETTE
13.
A.
(1)
(2)
(3)
VAN NUYS HIGH SCHOOL
What is the net force on mass #1?
FNET = m1a
FNET = (2.5 kg)(0.8 m/s2)
FNET = 2.0 N
14. What is the tension T?
14A. FNET = T = 2.0 N
15. What is mass #2, in kg?
15A.
(Fx: 0
(1)
(2)
(3)
(4)
(5)
(6)
(7)
Fy: T – w2)
Fy: FNET = ma = m(- a)  T – w2 = - m2a
T = w 2 – m2 a
T = m2g – m2a (because w2 = m2g)
T = m2 (g – a)
m2 = T / (g – a)
m2 = (2.0 N) / (9.8 m/s2 – 0.8 m/s2)
m2 = 0.22 kg
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