app of the 2nd law single body problems

Let’s recall:

1.How is the second law of motion represented mathematically?

2.What does the unit “newton” mean?

Application of the Second LawSingle-Body

Problems

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

PROBLEM-SOLVINGTECHNIQUES

Mixed Up Recipe

• 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.

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.

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

Draw a Free-Body Diagram

F = 4000 Nx

y

3

a

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 = ?

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 = ?


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

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

= +

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 = ?


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

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 = ?

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?


m Pf

1

2Given: a = 4.0 m/s2; m = 6.0 kg, f = 20.0 NFind: P = ?

a


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


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


Ta

1

2 Given: a = 5.0 m/s2; ` m = 100 kg

Find: T = ?

W

m


Ta

3

4 Upward (+)

+

W -

+

5 Fnet = T – W = T – mg


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

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

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?

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

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

Assignment• Devise your own problem-solving

strategy in solving problems in multiple-body systems in the application of the second law of motion?

app of the 2nd law single body problems

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