App of the 2nd law single body problems
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Let’s recall:
1.How is the second law of motion represented mathematically?
2.What does the unit “newton” mean?
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Application of the Second LawSingle-Body
Problems
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In which direction is the net force acting car A when it is moving east?
A
In which direction is the net force acting car B when it is braking to a stop while moving east?
B
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PROBLEM-SOLVINGTECHNIQUES
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Mixed Up Recipe
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• Read the problem carefully and then draw and label a rough sketch.
• Indicate all given information and state what is to be found.
• Construct a free-body diagram for the object undergoing acceleration and choose an x or y axis along the continuous line of motion.
• Indicate a consistent positive direction along the continuous line of motion.
• From the free-body diagram(s), determine the resultant force along the assumed positive line of motion, Fnet.
• Set the resultant force (Fnet) equal to the mass times the acceleration. Fnet = ma
• Substitute all given quantities and solve for the unknown.
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1. Read the problem carefully and then draw and label a rough sketch.
2. Indicate all given information and state what is to be found.
3. Construct a free-body diagram for the object undergoing acceleration and choose an x or y axis along the continuous line of motion.
4. Indicate a consistent positive direction along the continuous line of motion.
5. From the free-body diagram(s), determine the resultant force along the assumed positive line of motion, Fnet.
6. Set the resultant force (Fnet) equal to the mass times the acceleration. Fnet = ma
7. Substitute all given quantities and solve for the unknown.
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Illustrative Example no.11. A 1000-kg car travels on a straight highway with
a speed of 30.0 m/s. The driver sees a red light ahead and applies her brakes which exerts a constant braking force of 4.0kN. What is the acceleration of the car?
v = 30.0 m/s
a = ?
F = 4.0kN = 4000N
1
2
m = 1000 kg
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Draw a Free-Body Diagram
F = 4000 Nx
y
3
a
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Rightward Positive
F = - 4000 Nx
y
4
Fnet = - 4000 N5
Fnet = ma
a = Fnet
ma = -4000 N
1000 kga = -4.0 m/s2
6
7a = ?
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Illustrative Example no.22. A 15.0-kg box is pushed by two boys with forces
of 15.0 N and 18.0 toward the right. Neglecting friction, what is the magnitude and direction of the acceleration of the box?
1
2
m F1F2
F1 =15.0 NF2 =18.0 N
a Given: Find: a = ?
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Draw a Free-Body Diagram
Fnet = F1 + F2x
y
3
m = 15.0 kg
2. A 15.0-kg box is pushed by two boys with forces of 15.0 N and 18.0 toward the right. What is the magnitude and direction of the acceleration of the box?
a
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Rightward positive45
Fnet = ma
a = Fnet
ma = 33.0 N
15.0 kga = +2.20 m/s2
Fnet = F1 + F2
x
y
Fnet
Fnet = 15.0N + 18.0N = 33.0N
m = 15.0 kg
= + 6
7
a = ?
a
a = 2.20 m/s2, Right
= +
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Illustrative Example no.33. A freight elevator is
lifted upward with an acceleration of 2.5 m/s2. If the tension in the supporting cable is 9600 N, what is the mass of the elevator and its contents?
1
2
Ta
W
mGiven: a = 2.5 m/s2;
T = 9600 NFind: m = ?
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Draw a Free-Body Diagram
T
x
y
3
W
3. A freight elevator is lifted upward with an acceleration of 2.5 m/s2. If the tension in the supporting cable is 9600 N, what is the mass of the elevator and its contents?
a
Weight = mass x acceleration due to gravity
W = mg
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Upward, positive45
T = ma + mg = m(a + g)T
(a + g) m = 9600 N
2.5 m/s2 + 9.8 m/s2
m = 780 kg
Fnet = T – mg
6
7
T
x
y
W = - mg
T – mg = ma
m =
m = 9600 N12.3 m/s2
a
= +
= +
m = ?
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Solve the following
1. What horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?
2. A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?
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1. What horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?
m Pf
1
2Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 NFind: P = ?
a
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1. What horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?
Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 NFind: P = ?
3Pf a
4 Rightward ++
+-
5 Fnet = P – f
Free-body diagram
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1. What horizontal pull is required to drag a 6.0-kg box with an acceleration of 4.0 m/s2 if a friction force of 20.0 N opposes its motion?
Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 NFind: P = ?
5 Fnet = P – f
6 P – f = ma P = ma + f
7 P = (6.0 kg)( 4.0 m/s2)+ 20.0 N
P = 24.0 N + 20.0 N
P = 44.0 N Equate Fnet to ma, Fnet = ma
Derive equation to find the unknown
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2. A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?
Ta
1
2 Given: a = 5.0 m/s2; ` m = 100 kg
Find: T = ?
W
m
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2. A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?
Ta
3
4 Upward (+)
+
W -
+
5 Fnet = T – W = T – mg
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2. A 100-kg metal ball is lowered by means of a cable with a downward acceleration of 5.0 m/s2. What is the tension in the cable?
Ta
+
W -
+
5 Fnet = T – mg 6 T – mg = ma
T = mg + maT = m(g + a)
7 T = (100 kg)(9.8 m/s2+ 5.0 m/s2) T = 100kg(14.8 m/s2) = 1480 kg
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Solve in paper number 4.1. It is determined that the net force of 60.0 N will
give a cart an acceleration of 10 m/s2. What force is required to give the same cart an acceleration of (a) 2.0 m/s2? (b) 5.0 m/s2 and (c) 15.0 m/s2 ?Ans. (a)F = 12.0 N; (b) 30,0 N and F =90.0 N
2. A 20.0-kg mass hangs at the end of a rope. Find the acceleration of the mass if the tension in the rope is (a) 196 N; (b) 120 N; and (c) 260 N.
Ans. (a) 0; (b) -3.8 m/s2; (c) + 3.2 m/s2
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3. A 10-kg mass is lifted upward by light cable. What is the tension in the cable if the acceleration is (a) zero, (b) 6.0 m/s2 upward, and (c) 6.0 m/s2 downward?
4. An 800-kg elevator is lifted vertically by strong rope. Find the acceleration of the elevator if the rope tension is (a) 9000 N, (b) 7840 N, and (c) 2000 N?
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1. An unknown mass slides down an inclined plane. What is the acceleration in the absence of friction?
30°
xy
W
Wx
FN a m
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1. An unknown mass slides down an inclined plane. What is the acceleration in the absence of friction?
30°
xy
W
Wx
FN
aFnet = Wx = W sin 30° Fnet = mg sin 30 °
mg sin 30 ° = mag sin 30 °= a9.8 m/s2 sin 30 °= aa = 4.9 m/s2
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Assignment• Devise your own problem-solving
strategy in solving problems in multiple-body systems in the application of the second law of motion?