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# App of the 2nd law single body problems

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### Transcript

• 1. Let’s recall:
How is the second law of motion represented mathematically?
What does the unit “newton” mean?
• 2. Application of the Second Law
Single-Body Problems
• 3. 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
• 4. PROBLEM-SOLVING
TECHNIQUES
• 5. Mixed Up Recipe
• 6.
• Indicate all given information and state what is to be found.
• 7. Substitute all given quantities and solve for the unknown.
• 8. From the free-body diagram(s), determine the resultant force along the assumed positive line of motion, Fnet.
• 9. Set the resultant force (Fnet) equal to the mass times the acceleration. Fnet = ma
Read the problem carefully and then draw and label a rough sketch.
• Construct a free-body diagram for the object undergoing acceleration and choose an x or y axis along the continuous line of motion.
• 10. Indicate a consistent positive direction along the continuous line of motion.
• 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.
• 11. Illustrative Example no.1
1. 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?
m = 1000 kg
1
a = ?
v = 30.0 m/s
2
F = 4.0kN = 4000N
• 12. Draw a Free-Body Diagram
3
y
F = 4000 N
x
a
• 13. Rightward Positive
4
y
5
Fnet = - 4000 N
Fnet = ma
6
F = - 4000 N
Fnet
m
x
a =
-4000 N
1000 kg
a = ?
7
a =
a = -4.0 m/s2
• 14. Illustrative Example no.2
2. 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?
m
F1
1
F2
a
2
a = ?
Given:
Find:
F1 =15.0 N
F2 =18.0 N
• 15. Draw a Free-Body Diagram
3
y
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?
Fnet = F1 + F2
x
a
m = 15.0 kg
• 16. Rightward positive
4
5
Fnet = F1 + F2
y
Fnet = 15.0N + 18.0N = 33.0N
Fnet = ma
6
Fnet
= +
Fnet
m
a =
x
a
= +
m = 15.0 kg
33.0 N
15.0 kg
7
a =
a = ?
a = 2.20 m/s2, Right
a = +2.20 m/s2
• 17. Illustrative Example no.3
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?
1
T
a
m
Given: a = 2.5 m/s2; T = 9600 N
2
Find: m = ?
W
• 18. Draw a Free-Body Diagram
3
y
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?
T
a
x
W
Weight = mass x acceleration due to gravity
W = mg
• 19. Upward, positive
4
5
Fnet = T – mg
T – mg = ma
y
6
T = ma + mg
= m(a + g)
T
(a + g)
T
= +
m =
a
= +
9600 N
2.5 m/s2 + 9.8 m/s2
m =
7
x
m = ?
9600 N
12.3 m/s2
m =
W = - mg
m = 780 kg
• 20. Solve the following
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?
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?
• 21. 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?
a
1
P
m
f
2
Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 N
Find: P = ?
• 22. 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 N
Find: P = ?
+
a
3
f
-
Free-body diagram
P
+
4
Rightward +
5
Fnet = P – f
• 23. 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 N
Find: P = ?
5
Fnet = P – f
P = 24.0 N + 20.0 N
P – f = ma
6
Equate Fnet to ma, Fnet = ma
P = 44.0 N
P = ma + f
Derive equation to find the unknown
7
P = (6.0 kg)( 4.0 m/s2)+ 20.0 N
• 24. 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?
1
T
a
m
2
Given: a = 5.0 m/s2; ` m = 100 kg
W
Find: T = ?
• 25. 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?
3
T
+
a
+
W
-
4
Upward (+)
5
Fnet = T – W = T – mg
• 26. 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?
T
+
a
+
5
Fnet = T – mg
W
T – mg = ma
-
6
T = mg + ma
T = m(g + a)
7
T = (100 kg)(9.8 m/s2+ 5.0 m/s2)
T = 100kg(14.8 m/s2)
= 1480 kg
• 27. Solve in paper number 4.
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
• 28. 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?
• 29. An unknown mass slides down an inclined plane. What is the acceleration in the absence of friction?
y
FN
x
a
m
Wx
30°
W
• 30. An unknown mass slides down an inclined plane. What is the acceleration in the absence of friction?
y
FN
x
Fnet = Wx = W sin 30° Fnet= mg sin 30 °
a
mg sin 30 ° = ma
Wx
30°
g sin 30 °= a
9.8 m/s2sin 30 °= a
a = 4.9 m/s2
W
• 31. Assignment
Devise your own problem-solving strategy in solving problems in multiple-body systems in the application of the second law of motion?