Energy

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Energy  Work

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Energy is the ability to do work Definition of the work done by a constant force : If a constant force F acted upon a particle to transfer it form an initial position A to a new position B , this position is called the displacement of the particle is S , then the work done is the product of their algebraic measures

Rules of work

1 The work done by a horizontal

force

W  F S The work done by resistane W  -RS

 2  The work done by a force inclined with horizontal by angle  W  F S Cos 

 3  The work done by weight  motion is downward  a W  m g S The work done by weight

 motion is upward  b  W  - m g S  4 The work done by a force on an inclined plane

 If the motion is down   a  W  m g S Sin The work done by a force on an inclined plane

 If the motion is upward   b  W  - m g S Sin Dynamic – 3rd secondary

 - 942 -

Chapter Six – work - Energy

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 5 The work done by a force by given vector : W F

S  a1b1  a2b2 Or

F

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If F  a1 i  a2 j and S  b1 i  b2 j S  F S Cos   a1b1  a2b2

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Units of measure of work

Units of measure of work = unit of force magnitude  unit of length The important units are : (1) joule or (Newton – meter)

(2) Erg or (dyne – cm)

(3) Kilogram – meter (kgm.wt.m)

------------------------------------------------------------------------------------------------------The relation between units of measure of work  kgm.wt .m  9.8 joule  9.8 10 7 Erg 1 kg.wt.m 9.8 1  Erg  10 -7 joule  kgm.wt.m 9.8  107

 Joule  107 Erg 

------------------------------------------------------------------------------------------------------Example (1) A body of mass 200 kg moves in a st. line on a horizontal plane with uniform acceleration of magnitude 122.5 cm/sec 2 under the action of a horizontal force F and resistance of magnitude 12 kgm.wt starting from rest . calculate in kgm.wt.m the work done during 20 sec from the start by each of the following : (i)The force F . (ii)The resistance. (iii) The resultant force. (iv)The body weight . Answer 1 2 1 2 S  V t  at   S  0   122.5   20   24500 cm  S  245 metre 2 2 The equation of motion of the body is F  R  ma 1225  F  R  ma  12  9.8  200   9.8  12  25   9.8  37 Newton  9.8  37kgm.wt. 1000 F R First the work done by F  F  S  37  245  9065 kg.wt.m. Second the work done R  - R  S  -12  245  -2940 kgm.wt.m. 245 Third the work done F  R  ma  S  200   245  245  245 joule 200  245  245  9.8  6125 kgm.wt.m. Fourth the work done by the weight  0

Dynamic – 3rd secondary

mg

 mg  S  - 952 -

Chapter Six – work - Energy

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Example (2) A car moves along a horizontal road with uniform velocity of magnitude 9 km/h. Find the work done by the resistance force of the road during one minute from the start measure in kg.wt.meter given that the magnitude of the driving engine of the car equals 147 Newton. Answer There is a uniform velocity   F  R  147 Newton 5 V  9   2.5 m / sec , t  60 sec and S  V t Uniform velocity  18  S  2.5  60  150 m   W  -R S  -147  150  -22050 joules   9.8   -2250 kg.wt.m

-------------------------------------------------------------------------------------------------------

Example (3) A car of mass 1.5 ton, moves along a horizontal straight road, its driving engine force is stopped and its brakes are used when its velocity is 63 km/hr. So, it stopped after covering 12.5 m from the starting instant. Find the work done by the brakes resistance during this distance assuming that the magnitude of the resistance is constant. Answer Engine stopped means that F  0 , Brakes are used means that acceleration is negative

Velocity is stopped means that V  0 

 - R  m a     1

5  49   0   63    2 a  12.5   a  - m / sec 2 18  4  49 Substitute in  1 : - R  1.5  1000   R  18375 Newton   9.8   1875 kg.wt 4  W  -R S  1875  12.5  -23437.5 kg.wt.m And

V 2  u2  2 a s

------------------------------------------------------------------------------------------------------Example (4) 2 A body of mass 2 kg is projected up a smooth plane inclined to the horizontal at 30 o with 7 velocity 14.7 m / sec . Calculate in units of joule the work done by its weight during its maximum displacement on the plane .

Answer 1  -4.9 m / sec 2 2 14.7  14.7 441 S  m 2  4.9 20

F, R are neglected  a  - g Sin θ  9.8Sin30 o  -9.8Sin 30 o  -9.8  V 2  U 2  2a s   0  14.7   2  4.9S 2

 W  -m g s Sinθ  -



16 441 1  9.8    -246.96 joule 7 20 2

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Dynamic – 3rd secondary

- 952 -

Chapter Six – work - Energy

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Example (5) A body of mass 15 kgm moves along the line of greatest slope of a plane inclined at an angle 3 whose Sine  with the horizontal under a force F acting along the line of greatest slope up 5 the plane . If the body moves with uniform acceleration of magnitude 4.9 m/sec 2 starting from rest. Calculate the work done by F and by the weight of this body during an interval of magnitude 8 sec from start. Answer 2 a  4.9 m / sec u 0 W  ?? t  8 sec F S 1 1 2 U S  u t  at 2  S  0   4.9  8   156.8 m 2 2 As  S  is from the starting to the end of the inclined plane m g sin   m g cos  So we will use the rule :



F  mg sin  m a

mg 3  F  m g sin  m a  15  9.8   15  4.9  4.9  18  15   4.9  33 Newton   9.8   16.5 kgm.wt 5  W  F  S  16.5  156.8  2587.2 kg.wt.m  The work done by the weight  - m g S Sin  15  9.8  156.8  -

3 joules   9.8   -1411.2 kgm.wt.m 5

------------------------------------------------------------------------------------------------------Example (6) A body of mass 4 kg is placed on a plane inclined to the horizontal at an angle 30 o . A force 40 3 Newton lying in the vertical plane passing through the line of greatest slope and inclined at an angle 30° upwards acts on the body and makes it move 3 meters up the plane. Find in units of joule :

 a  The work done by the force during this displacement .  b  The work done by the weight during the same displacement. 40 3

Answer 3  3  180 joule 2 Important Note : we didn't use F  m g Sin  ma because  S  started from

The work done by the force  F Cos 30 o  S  40 3 

S

40 3 Cos 30o

30 o 3m

the action of force not from the begining of the plane 1 m g Sin 30o o 30 2  -6 kg.wt.m  -6  9.8 joule  -58.8 joule

The work done by the weight  - m g s Sin30 o  -4  3 

30 o

m g Cos 30o

mg

------------------------------------------------------------------------------------------------------Dynamic – 3rd secondary

- 959 -

Chapter Six – work - Energy

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Example (7) A body of mass 200 gm. is projected vertically upward s with velocity 9.8 m / sec . Find the work done by its weight during the interval of the first three seconds from the instant of projection measured in units of Erg. Answer 1 2 1 S  ut  gt   S  9.8  3   9.8  9  S  -9.8  1.5  -14.7 m 2 2  W  - mg  -S  m g s  W  200  980  1470 erg  28812  10 4 erg

------------------------------------------------------------------------------------------------------Example (8) A bullet of mass 25 gm. is fired horizontally at a fixed target with velocity 5.6 m/sec and 1 sec. Calculate the work done by the resistance 35 force during this interval assuming that it is constant. Answer 1 1 u  5.6 m / sec t  sec V  0  and V  u  at  0  5.6  a 35 35 25  a  -196 m/sec 2 and - R  ma    -196   -4.9 Newton 1000

embedded through it and comes to rest after

1 S  u t  at 2  2

2

1 1  1   S   5.6    -196     0.08 m 35 2  35 

W  - R S   W  -4.9  0.08  -0.392 joules  107   -3920000 Erg

------------------------------------------------------------------------------------------------------Example (9) A force F  21i  28 j acts on a body so that it moves from point A   2,-1 to point B   -1,3

where i and j are two perpendicular unit vectors in the direction of OX ,OY .The magnitude of F is measured in Newton and the magnitude of the displacement is measured in metre. Prove that the force and the displacement vectors are parallel, then find the work done in units of kg.wt. Answer -4 S  AB  B  A   -i  3 j    2i  j   -3i  4 j  its slope is 3 -4 F  21i  28 j  its slope is 3  F and S are parallel and in opposite directions. W  F W -

S   21i  28 j 

 -3i  4 j   -63  112  -175 Newton.m.

175 125 kg.wt.m 9.8 7

Dynamic – 3rd secondary

- 952 -

Chapter Six – work - Energy

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Example (10) If i and j are two unit vectors in a cartisian system of 2 perpendicular axes and a force F  -3i  7 j acted upon a particle to move it from a point A   -2,-1 to a point B   4,1 in a straight line and move under the same force in a st - line from B to a third point C   1,5  . Prove that the sum of work done by this force during these 2 displacements equals the work done by the same force during thier resultant displacment.

Answer S1  AB  B  A  4i  j  -2i  j  6i  2 j

    S  BC  C  B   i  5 j    4i  j   -3i  4 j S  AC  C  A   i  5 j    -2i  j   3i  6 j The work done by F during S  F S   -3i  7 j  6i  2 j  2

1

1

 w1  -3  6  7  2  -4 units . The work done by the F during S2  F



S 2  -3i  7 j

 w2  -3  -3  7  4  37 units . The work done by the F during S  F  w  -3  3  7  6  33 units



S  -3i  7 j

  -3i  4 j 

  3i  6 j 

  w1  w2  -4  37  33  w

------------------------------------------------------------------------------------------------------Example (11) A force F  5i  10 j acts on a body , and the displacement of the body at any instant  t  is

given by the relation S   t  2  i   t 2  1 j where i and j are two perpendicular unit vectors in the directions of OX , OY . and the magnitude of F is measured in units of gm.wt. and the the unit of the displacement is in metre , calculate in units of Erg the work done by the force between the two instants t  1 and t  4. Answer when t  1   S1  3i  2 j & when t  4  S2  6i  17 j S The displacement between t  1 and t  4   S2  S1  S   6i  17 j    3i  2 j   3i  15 j  W F

S   5i  10i 

 3i  15 j   15  150  165 gm.wt.m.

W  165  980  100  1617  10 4 erg

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Dynamic – 3rd secondary

- 954 -

Chapter Six – work - Energy

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The Power

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Power is how fast the work is done …. so P= W/t

 1

Definition of the power:

" Power is the time rate of doing work" Or "Power is the work done in unit of time" If the work done by force F during a time T 

 Power 

dw dt

 2  Rule of power in the linear motion under a constant force parallel to the line of motion : The force F and the displacement S in the same direction  work  F  S , where F , S are the algebaric measures of F and S d F  S dw ds   F   F V   Power  F  V dt dt dt Note : we can determine Power at any instant , while we can find Work between two

 Power 

instants only or during a given displacement

------------------------------------------------------------------------------------------------------Power  R  ma V * When the motion is horizontal with maximum velocity : F  R  0 * When the motion is horizontal : F  R  m a  we can say :

we can say :

Power  R 0 V

* When the motion is upward : F  m g  m a  we can say :

Power  m g  ma V

* When the motion is upward with maximum velocity : F  m g  0  we can say :

Power m V

* When the motion is on inclined plane to the horizontal upward : F  R  m g Sin  m a We can say :

P  R  m g Sin  m a V

* When the motion is on inclined plane to the horizontal upward with maximum velocity:  F  R  m g Sin  0  we can say :

P  R  m g Sin  0 V

------------------------------------------------------------------------------------------------------The Important Units are:

1 Watt



 Joul / sec

 Newton . Meter / sec 

 3  Horse Power  HP   Dynamic – 3rd secondary

 2  Erg/sec : dyne . cm /sec 

735 Watt  735  107 Erg / sec - 955 -

Chapter Six – work - Energy

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Example (1) A man of mass 75 kg ascends on a vertical ladder with uniform velocity 42 cm/sec. Calculate the Power of the man in horse unit. Answer F  F  mg  0   F  m g  75  9.8  735 Newton V  42 cm / sec 

42  0.42 meter / sec 100

 The man's power  F  V  735  0.42  308.7 watt 

mg

308.7  0.42 HP 735

------------------------------------------------------------------------------------------------------Example (2) A train of mass 300 tons and the power of its driving force  480 horse is moving on horizontal straight road with its maximum velocity which is 54 km / h. Find magnitude of the road resistance per each ton of train mass.

Answer The train is moving with uniform velocity :  F  R  0   F  R 5 Power  F  V   480  735  R  54    R  23520 Newton 18  Resistance per ton  23520  300  78.4 Newton   9.8   8 kgm.wt

------------------------------------------------------------------------------------------------------Example (3) A train of mass 98 ton and the power of its engine 120 horse moves along a straight horizontal road with maximum velocity 162 km / h. Find magnitude of maximum velocity with which that 1 train can ascend a straight road inclined at angle of sine with the horizontal given that 140 magnitude of resistance of the two roads are equal. Answer 5 In the motion on the horizontal road: Power  F  V   120  735  F  162  18  F  1960 Newton  The train is moving with maximum velocity

 The train motion is uniform



 R  F  1960 Newton

R

F

In the motion on the inclined road:  F  m g sin  R  0 And 

Power never change 

P 1  98  1000  9.8   1960  0 V 140

120  735  8820  V  10 m / sec V

F

mg sin 

R



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Dynamic – 3rd secondary

- 952 -

Chapter Six – work - Energy

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Example (4) A car of mass 6 tons and the power of its engine is 30 horses is moving along a horizontal road against a constant resistance with maximum velocity of magnitude 108 km/h . Find the magnitude of this resistance per each ton of the car mass, if this car pulled another car having a broken motor 1 and with mass 2 tons by means of a rope along a road inclined at angle of Sin to the horizontal. 100 Find the magnitude of the maximum velocity with which the two cars can move up this inclined road given that the power of the first car and resistance per ton are not changed . Answer 5 P  30  735  22050 watt V1  108   30 m / se 18 On horizontal road : The motion is uniform   F  R where F1 

P 22050   735 Newton   R  735 Newton V1 30

Then the resistance per each ton :

735  122.5 Newton 6

On inclined road : Mass  6  2  8ton  8000 kg And the resistance per each ton  122.5  8  980 The motion is uniform   F2  m g sin  R2  0 and the  F2 

power is unchanged

P 22050 22050 22050  1      8000  9.8    1764   980  0  V2 V2 V2 V2  100   V2  12.5 m / s

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Dynamic – 3rd secondary

- 952 -

Chapter Six – work - Energy

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Example (5) 1 A car of mass 15 tons moves upwards along a road inclined at an angle of sin with the 50 horizontal with maximum velocity 72 km/hr , and if the car moves downwards on the same road with maximum velocity of magnitude 108 km/hr , find the power of the car engine given that the magnitude of the road resistance is proportional to the magnitude of the car velocity . Answer 5  first  During the ascend of the car: V1  72   20 m / sec 18 P P The motion is uniform   F1  m g sin  R1  0  where F1   V1 20 1 1  1   P  15  1000  9.8     R1  0   P  2940  R1  0     1 F 20 50 20    Second  During the descend of the car: Power never change R mg sin  5 V2  108   30 m / sec  18 P P The motion is uniform   F2  m g sin  R2  0 where F2   V2 30 1 1  1   P  15  1000  9.8     R2  0   P  2940  R2  0  30  R 30 50 30    P  88200  30R2  0     2  F2 mg sin  R V 20 2 And R V   1  1    R1  R2  R2 V2 30 3 1 2 40 Then substitute in  1 :  P  2940  R2  0  20    P  58800  R2  0     3  20 3 3 50 50 Then subtract  2    3  : 147000  R2  0   R2  147000  R2  8820 Newton 3 3 Then substitute in  2  :  P  88200  30  8820   0  P  176400 Watt   735   240 HP

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

- 952 -

Chapter Six – work - Energy

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Example (6) A train of mass  m  tons moves upwards along the line of the greatest slope of a plane inclined

1 to the horizontal with maximum velocity 36 km/hr against a resistance of 125 magnitude 16 kg.wt per each ton of the train mass, after reaching the top of the inclined plane, the at an angle of sin

last wagon of mass 15 ton is separated from the train, then the train is moved once more downwards along the inclined road with maximum velocity 120 km/hr , given that the magnitude of the resistance per ton is the same in the two cases, calculate the value of  m  and the power of the train engine . Answer 5  first  During the ascend of the car: M train  1000 m V1  36   10 m / sec 18 P P Resistance  16  9.8  m  156.8 m and F1   V1 10 F The motion is uniform   F1  m g sin  R1  0 R 1  1  mg sin   P  m  1000  9.8    156.8 m  0  10  125   1 1  P  78.4 m  156.8 m  0  P  235.2 m  0 10 10  P  2352 m     1

 Second  During the descend of the car:

M train   m  15  1000 

5 100  m / sec and Resistance  16  9.8  m  15  18 3 The motion is uniform  Power never change 

V2  120 

R

mg sin 

P 3P F2  m g sin  R2  0 where F2   V2 100

F2



3  1   P   m  15  1000  9.8     156.8  m  15   0 100 125   3 3  P  78.4 m  1176  156.8 m  2352  0   P  78.4 m  1176  0  100  100 100  3P  7840 m  117600  0     2 

Substitute  1 in  2   3  2352m   7840 m  117600  0  7056m  7840 m  117600  0

 784 m  117600   m  150 ton Then substitute in  1 :  P  2352  150  352800 Watt   735   480 HP

Dynamic – 3rd secondary

- 952 -

Chapter Six – work - Energy

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Example (7) If i and j are the unit vectors parallel to the perpendicular axis of a cartisian system and constant force F  4i  3 j acted upon practical to make a displacement given by : 1  S   2t  1 i   t 2  t  j where t is in sec and magnitude of F in Newton, S in meters 2  Calculate  First  the work done by F from t  2 sec to t  6 sec

 Second 

the power of this force when t  3 sec

Answer The work done by F during an interval  t   F S  4i  3 j





 1 2  2t  1 i     t t  2  

 j 

3 1   W  4  2t  1  3  t 2  t   W  t 2  5t  4 2 2  3 2  The work done during 2 sec  w  2    2   5  2  4  20 Joule 2 3 2  The work done during 6 sec  w  6    6   5  6  4  88 Joule 2  The work done during the interval form 2 to 6 sec  w  6   w  2   88  20  68 Joule dw the power   3t  5 so when t  3 sec , the power  3  3  5  14watt dt ------------------------------------------------------------------------------------------------------Example (8) If i and j are the unit vectors parallel to the perpendicular axis of a cartisian system and constant force F  ki  b j acted upon a particle to make a displacement given by : S  4  t 2  3t  2  i  3t 2 j where t is in sec and magnitude of F in dyne, S in cm, and the Work done by F during the first 3 seconds is 4780 Erg and the power of F when t  2 sec is 1520 Erg/sec, then find the magnitude of F .

Answer The work done by F during an interval  t   F S  k i  b j





4  t 2  3t  2  i  3t 2 j   

 W  4k  t 2  3t  2   3bt 2  W  4 k t 2  12 k t  8 k  3bt 2

The work done during 3 sec  4780 Erg  36k  36k  8k  27b  4780 dw  80 k  27 b  4780     1 And the power   8k t  12k  6bt dt So when t  2 sec and P  1520   16k  12k  12b  1520  28k  12b  1520     2  from  1 and  2  :

80 k  27 b  4780 28k  12b  1520

 -28   80 

 -2240  756 b  -133840  2240  960b  121600

 after addition : -204b  -12240   b  60 Then Then F  80i  60 j Dynamic – 3rd secondary

 its magnitude is F 

 80 

- 922 -

k  80 2

  60   100 dyne 2

Chapter Six – work - Energy

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Energy

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We have two types of energy

Potential Energy

Kinetic Energy

The energy that is stored in an object

The energy which is moved in an object

‫الطاقة الثابتة‬

‫الطاقة المتحركة‬

Rule

P m g h

Rule

h : height from the refrence point on the ground

K .E.  T 

1 mV 2 2

V : Velocity done by the body

 the lowest possible point of the body  Depends on Motion

Depends on Position When vector exists : PA  F

AO



1 When vector exists: T  m V 2

Definition The Potential energy of a particle denoted by P at a certain instant is defined as the work done by acting forces on the body if it moved it from its position at this instant to a fixed posiion on the straight line on which motion occurs .

units



Definition The Kinetic energy of a body denoted by T is defined as the product of half the mass of the particle times the square of the magnitude of its velocity

Joules  Newton.m

Joules  Newton.m

units

Erg  dyne.cm

Dynamic – 3rd secondary

V

- 922 -

Erg  dyne.cm Chapter Six – work - Energy

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for more understanding

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high P.E.  m g hbig Zero K.E.  u  0 

Meduim P.E.  K.E. low P.E.  m g hsmall Zero K.E. V  0 

high K.E.

Reference point

-------------------------------------------------------------------------------------------------------

Pendulum P.E.  m g h1

P.E.  m g h3

K.E.  0 as V final  0

K.E.  0

h1

h3

h2 P.E.  K.E. K.E. exists only

Reference point

-------------------------------------------------------------------------------------------------------

Relation between Work and Energy

T  To  Wnet

P  Po  -W

1 m V 2  Vo 2   Fnet  S 2

m g  h  ho   -F  S Change P.E. is the work done

Change K.E. is the net of the work done

 To  Po  T  P Dynamic – 3rd secondary

- 929 -

Chapter Six – work - Energy

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The proof of the rule : T  To  W

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The change of Kinetic energy of a particle when it moves from an initial position to a final position is equal to the work done by the force acting on it during displacement between these 2 positions T 

1 dT 1 dv mv 2    m  2v  mv  a   m a   v 2 dt 2 dt

dT  F  V     1  dt

The Power  F  V  

dw  F V    2 dt

dT dw d  0  T  W   0 dt dt dt  T  W  K Where k is constant  but W  0 when T  T0  To  0  K  K  To Substracting  1   2  :

 T  W  T0





T  T0  W

Then Change of K .E energy  Work done or T  T0  W -------------------------------------------------------------------------------------------------------

Very important notes :

1 Kinetic energy must be Positive  2  To find the Kinetic energy lost  T1  T2  3  To find the Kinetic energy gained  4  When P.E. is maximum  K.E. is Zero  as hmax leads to V  0   5  When K.E. is maximum  P.E. is Zero

 T2  T1

------------------------------------------------------------------------------------------------------Example (1) Two smooth spheres move in a horizontal straight line in two opposite directions, the mass of the first is 5 kgm, the 2 balls get in impact, just before impact the velocity if the first is 30 cm/sec and the second is 50 cm/sec and just after impact the first rebounds with velocity 10 cm/sec and the second comes to rest, find the mass of the second sphere and the K.E lost due to collision. Answer  To find m : m1V1  m2V2  m1V '1  m2V '2  5  30  m  50  -5  10  m  0  m  4 kgm To find K .E lost due to collision :

before impact : v1  30

I

5 kgm

v2  50

m kgm

I

Sum of K .E of the 2 spheres before collision : After impact :

v1  10 v2  0 1 1 2 2   5   0.30    4  0.5   0.725 joule 2 2 1 2 Sum of K .E of the 2 spheres after collosion :  5   0.1  Zero  0.025 joule   2 And K .E lost  Sum before  Sum after  0.725  0.025  0.7 joule Dynamic – 3rd secondary

- 922 -

Chapter Six – work - Energy

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Example (2) A body of mass 4 kgm moves from rest along a horizontal straight line a distance 5 meter under a horizontal force of magnitude 180 gm.wt , find

 First  the work done by this force during this distance  Second  the kinetic energy gained at the end of this distance Third  the magnitude of the final velocity . Answer First : The work done: W  F  S  180  980  500  882  10 5 Erg  882  10 5  107   8.82 Joule

Second : K .E gained : T  T0  work done  8.82 Joule 1 1 1 1 Third : To find V : T  T0  mV 2  mVo 2    4  V 2  m  zero  8.82 2 2 2 2 2 V  4.41  V  2.1 meter / sec

------------------------------------------------------------------------------------------------------Example (3) A body of mass 2.5 kg fell vertically from a point A above the ground surface to reach a point B on the ground surface where its kinetic energy is 245 joule , find the potential energy of this body at A in kg.wt.m , then calculate its kinetic energy and the magnitude of its velocity when it is at a height 6.4 m above the ground surface . Answer A Always assume a reference level which is the lowest point TA  0 TA  PA  TB  PB   0  PA  245  0 PA  m g hA  PA  245 joules   9.8   25 kg.wt.m

When the body is 6.4 m above the ground TC  PC  TB  PB Where PC  m g hC  2.5  9.8  6.4   156.8 joules  TC  156.8  245  0  TC  88.2 joules 1 1  mVC 2  88.2    2.5 VC 2  88.2 2 2 2  VC  70.56  VC  8.4 m / s

hA

6.4

TC  ?? PC  m g hC

PB  0  as h  0  TB  245 joules B Reference level [

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

- 924 -

Chapter Six – work - Energy

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Example (4) A body starts motion from the top of an inclined plane to its bottom along the line of the greatest slope, and starting from rest, if the potential energy at the top of the plane equals 17.64 joule and the magnitude of its velocity when it reaches the bottom of the plane is 8.4 m/s Then find the mass of this body and the height of the plane . PA  17.64 Answer TA  0  u  0  Always assume a reference level which is the lowest point When the body is at A : PA  m g hA  17.64  m  9.8  hA A

 m hA  1.8     1

hA

PB  0 TB  ???

When the body moved from A to B TA  PA  TB  PB  0  17.64  TB  0 TB  17.64 joules 

B

Reference level

1 1 1 1.8 2 mVB 2  17.64   m  8.4   17.64   m  kg then from  1 hA   3.6 m 2 2 2 0.5

------------------------------------------------------------------------------------------------------Example (5) A body of mass 50 kg is left to fall from height 40 m above the ground surface, calculate the kinetic energy when it reaches the ground surface, and if this body embedded a distance 80 cm through the ground surface before coming to rest, prove that the magnitude of the ground resistance equals 2550 kg.wt assuming that it is constant . Answer

u 0

V  u  2 g S   V  2  9.8  40   784  V  28 m / s 2

2

2

1 1  T  mV 2   50 784   19600 joule 2 2 To get the resistance inside the sand , we have two methods

40

0.8

R

T1  ??

sand

a mg

T2  0

st

1 method m g  R  m a where V 0 u  28 S  0.8 And

V 2  u2  2 a S

 0   28   2  0.8  a 2

 a  -490 m / sec 2  50  0.8   R  50  -490   R  24990 Newton  2550 kg.wt

2 nd method " I prefer this method when K.E. is given in the problem"  no acceleration is given 

T2  T1  Wnet  Fnet  S   m g  R  S   0  19600   50  9.8   R  0.8 

-19600  - 24500 0.8  R  24990 Newton  2550 kg.wt  50  9.8   R 

------------------------------------------------------------------------------------------------------Dynamic – 3rd secondary

- 925 -

Chapter Six – work - Energy

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Example (6) A car of mass 2.45 tons moves along a horizontal st. road with uniform acceleration, if its velocity at a certain instant was 27 km/hr and after covering a distance 9 m , its velocity became 81 km/hr , find the kinetic energy gained during this distance, and the magnitude of the force of the car motor given that the magnitude of the resistance to the car motion equals 80 kg.wt per ton . Answer 5 m  2.45  1000  2450 kg V1  27   7.5 m / s S 9 m 18 5 V2  81   22.5 m / s F  ??? R  80 kg per each ton 18 1 1 2 2 K.E. gained  T2  T1  m V2 2  V12    2450   22.5   7.5   551250 joule 2 2  56250 kg.wt.m To get the force of the car motor , we have two methods :



1st method F  R  m a where R  80  9.8  2.45  1920.8 Newton V 2  u2  2 a S

And

  22.5   7.5   2  a  9  2

2

 a  25 m / sec 2  F  1920.8   2450  25 



2 nd method " I prefer this method when K.E. is given in the problem"  no acceleration is given 

T2  T1  Wnet  Fnet  S

 551250   F  R  S  551250   F  1920.8  9  551250  61250 9  F  63170.8 Newton  6446 kg.wt  F  1920.8 

 F  63170.8 Newton  6446 kg.wt

------------------------------------------------------------------------------------------------------Example (7) A car of mass 2 tons, when its velocity was 7 m/sec , its engine stopped and it continued a motion in a straight line under resistance of magnitude 250 kg.wt and it stopped. find K.E lost and the distance it travelled during this motion Answer 1 1 1 2 K .E. lost : T1  T2  mV12  mV2 2   2000 7   0  49000 Joule 2 2 2

T1  T2  -W  -  -R  S   49000  250  9.8 S



 S  20 meter

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

- 922 -

Chapter Six – work - Energy

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Example (8) If i and j are the unit vectors parallel to the perpendicular axis of a cartisian system and a force F  9i  12 j acted upon a particle , if the position vector of this particle as a function of time t in seconds is r   t 2  5  ˆi   t 2  1 ˆj , and the particle moved from A to another point B during 3 seconds , find :  i  The potential energy of the particle at A  ii  The change in the potential energy during this interval Given that the magnitude of F in Newton and the magnitude of r in meter. Answer 2 2 F   9 ,-12  and r  t  5 ˆi  t  1 ˆj



 



Point A means when t  0  ro  5iˆ  ˆj   5 ,-1 and point B means when t  3  r  14iˆ  8 ˆj   14 ,8  So  i  Potential energy at A : PA  F  PA   9 ,-12 

AO , where AO  O  A   -5 ,1

 -5 ,1  - 45  12  -57

joules

 i  Change in P.E.  PA  PB  -W  - F S , where S  AB  B  A   9 ,9   PA  PB   9 ,-12   9 ,9   -  81  108   27 joules

------------------------------------------------------------------------------------------------------Example (9) If i and j are the unit vectors parallel to the perpendicular axis of a cartisian system and a force F  8i  3 j acted upon a particle , if the change in kinetic energy of this particle when it moves from point A to point B  5iˆ  4 ˆj equals 34 joule, calculate the potential energy at A given that  0 ,0  is the zero point of the potential energy and the magnitude of F in Newton and the displacement in meter. Answer F   8 ,-3  , B   5 ,-4  , T  To  W  34 joule

W F So

S , where S  AB , so let the posotopn vector of A be  x , y 

S  AB  B  A   5 ,-4    x , y    5  x , -4  y   W F

S   8 ,-3 

 5  x , -4  y   34

40  8x  12  3 y  34  -8 x  3 y  -18     1 To get the Potential energy at A : PA  F  PA   8 ,-3 

AO , where AO  O  A   - x ,- y 

 - x ,- y   -8 x  3 y     2 

, then from  1 and  2  :  PA  -18 joule

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

- 922 -

Chapter Six – work - Energy

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Example (10) A cyclist is moving on a straight horizontal road starting from rest with uniform acceleration and used his legs in moving the bicycle to cover a distance of 320 m, where the kinetic energy of the bicycle and the cyclist together became 1600 kg.wt, at this instant the cyclist stopped moving his legs to cover another 150 m, where their kinetic energy together became 1150 kg.wt Calculate the magnitude of each of the driving force of the bicycle and the resistance of the road assuming that it is constant . Answer TC  1150 TB  1600 u 0 K.E. is given in the problem , so we R R F will solve the problem by using: C A B 320 m 150 m T2  T1  Wnet  Fnet  S Motion from A  B

TA  0 , TB  1600  9.8  15680 joule , Fnet  F  R , S  320 m

TB  TA  Wnet  Fnet  S  15680   F  R  320    F  R  49     1 Motion from B  C

TB  15680 joule

, TC  1150  9.8  11270 joule

,

S  150 m

TC  TB  Wnet  Fnet  S  11270  15680  - R  150    R  29.4 N   9.8   3 kg.wt Then substitute in  1 :

F  29.4  49   F  78.4 Newton   9.8   8 kg.wt

------------------------------------------------------------------------------------------------------Example (11) A tram wagon in the state of rest is pulled by a horizontal rope inclined at an angle of measure 60 o to the direction of the tram rails so it moved a distance 24.5 m against a resistance of magnitude 3000 kg.wt and the kinetic energy gained during this distance equals 18375 kg.wt.m calculate :  i  The work done by the resistance

 ii  The work done by the tension in the rope

 iii  The magnitude of the tension force in the rope . S  24.5 m

R  3000  9.8  29400 N

Answer T2  T1  18375  9.8  180075 joule

T

 i W  - R S  -29400  24.5  -720300 joule   9.8   -73500 kg.wt.m  ii  W  T Cos60 o S

60o

R

we have two unknowns, so we will use the rule of K.E. T2  T1  Fnet  S  T Cos60  R   S    T2  T1  T Cos60  R   S  o

T Cos60 o

24.5 m

o

 180075  T S Cos60 o  R S   180075  T S Cos60 o  720300 Work done by the tension in the rope  W  T S Cos60 o  900375 joule   9.8   91875 kg.wt.m

 iii Then the tension force : Dynamic – 3rd secondary

T S Cos60 o  900375  T 

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900375  73500 joule  24.5   Cos60 o  Chapter Six – work - Energy

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Example (12) A body of mass 4 kg is projected from the bottom A of an inclined plane towrds another point

B on the plane with velocity of magnitude 7 m/s along the line of the greatest slope AB . If the kinetic energy of this body at a point C bisecting AB equals 80.36 joule , and A lies on the ground surface , find :  first  the potential energy of the body at C and its height

 second  Prove that the change in potential energy from C to B equals the change in kinetic energy in the same distance  -17.64 Answer 1 2 When the body moved from A to C TA  PA  TC  PC   4 7   0  80.36  PC 2 17.64 B  PC  17.64 joules  m g hC  17.64  hC   0.45 m  4  9.8  C To find the change in potential energy from C to B

PC  PB  17.64  m g hB where hB  2hC  0.9 m  PC  PB  17.64   4  9.8  0.9   -17.64 joule To find the change in kinetic energy from C to B

hB

hC

A

Reference level

then we have to get the velocity at B 1 2 TA  PA  TB  PB   4 7   0  TB  4  9.8 0.9  2 TB  62.72 joules   TC  TB  62.72  80.36  -17.64 joule

------------------------------------------------------------------------------------------------------Example (13) A simple pendulum consists of a light string of length 180 cm carrying at its end a body of

mass 400 gm and move through an angle of measure 120 o , find :  i  The potential energy lost due to the motion of the body from the initial point of the path of motion to the mid point of the path  ii  The magnitude of the velocity of the body at the mid point of the path Answer At the mid point of the path , then   60 o   hB  0.9 m And PB  m g hB  0.4  9.8  0.9  3.528 joule At the mid point of the path   K .E.B  P.E.B 1  K.E.  mVB 2  3.528  VB  4.2 m / sec 2

Dynamic – 3rd secondary

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1.8 m

60 o o 60

0.9 m

30 o

B

hB

0.9 m

A 0.4

Chapter Six – work - Energy

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Example (14) When building the foundation of a house, a hammer of mass 210 kg is used to fall down from height 90 cm on a cylinder of mass 140 kg to push it in the ground a distance 18 cm, find :

 i  The magnitude of the common velocity of the cylinder and the hammer just after impact  ii  The lost in kinetic energy due to impact  iii  The magnitude of the average resistance of the ground to the cylinder in kg.wt Answer  i  To get the velocity of the hammer before collision : u 0

V1  ??

S  0.9 and

2

V12  2  9.8  0.9   17.64   V1  4.2 m / s So

u 0

V1  u  2 g S 2

0.9

R V1

m1V1  m2V2  V '  m1  m2 

 210  4.2   140  0    210  140 V '   V '  2.52 m / s 0.18

 ii The kinetic energy before impact :

V' mg

T1 T2 T3  0

V2  0 1 1 2 mV12   210  4.2   1852.2 joule 2 2 1 1 2 The kinetic energy after impact : T2  mV ' 2   350  2.52   1111.32 joule 2 2 Then the lost in kinetic energy  T1  T2  1852.2  1111.32  740.88 joule

T1 

 iii  To find R , it is easy to use K.E. rule : T3  T2  Wnet  Fnet  S   m g  R  S 

  0  1111.32   350  9.8   R  0.18 

-1111.32  -6174  R  9604 Newton  980 kg.wt 0.18 ------------------------------------------------------------------------------------------------------ 350  9.8   R 

Dynamic – 3rd secondary

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Chapter Six – work - Energy

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Example (15) A body of mass 20 kgm is projected with velocity 2.1 m/sec , from the top of an inclined plane in the direction of its line of greatest slope downwards to reach its bottom with velocity 5.6 m/sec , Find the work done by the plane resistance given that the height of the plane  1.5 m , and if Sin the angle of inclination of the plane with the horizontal is in Newton. V1  2.1 m / sec

3 . Then find magnitude of the resistance 14

Answer V2  5.6 m / sec



R

Assume that the distance from the top of the plane to its bottom is S. 1.5 3 1.5 1.5 m  m g Sin  Sin      S 7m S 14 S we can apply rule of energy and work during S  1 So T2  T1  Wnet  m V2 2  V12    m g Sin   R  S 2 1 3 2 2     20  5.6    2.1    20  9.8   R   S     2 14  - 24.5  269.5  42  S   R S  - R S  269.5  294  - 24.5  R  3.5 Newton -7

------------------------------------------------------------------------------------------------------Example (16) A body is planed at the top of an inclined plane of length 18 m inclined at an angle of measure 30 to the horizontal and is left to slide along the line of greatest slope. If its kinetic energy at the bottom of the incline is 4.41 kg.wt.m and the body continues its motion on horizontal plane under the same resistance of the inclined plane, and comes to rest after covering 50 cm on the 3 horizontal plane. Given that the body looses of its kinetic energy when it moves from the 4 inclined plane to the horizontal plane. Find magnitude of the resistance and the mass of the body. Answer u  0  T1  0 On the inclined : T2  T1  Fnet  S   T2  T1   mg sin  r   S R  4.41  9.8  0   4.9 m  R   18     1 m g Sin30

On the horizontal : T4  T3  Fnet  S   T4  T3  -R  S

1 1 0  T2  -R  S   -  4.41  9.8  -R  0.5 4 4  R  2.205  9.8 Newton , From 1   m  4.9 kg

R

T2  4.41

T4  0 0.5 m 1 T3   4.41 4

------------------------------------------------------------------------------------------------------Dynamic – 3rd secondary

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Chapter Six – work - Energy

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Example (17) A body starts motion from the top of an inclined plane to its bottom along the line of the greatest slope, and starting from rest, if the potential energy at the top of the plane equals 17.64 joule and the magnitude of its velocity when it reaches the bottom of the plane is 8.4 m/s Then find the mass of this body and the height of the plane . Answer TB  PA  1.225  9.8  12.005 joules and W  -3.675  9.8  -36.015 joules TB  12.005  i  TB  TA  W   12.005  TA  -36.015 TA  48.02 joules

 ii 

B

PB  PA  -W   PB  12.005  36.015  PB  48.02 joules

t  2 sec

PA  12.005

1 mVB 2  12.005     1 A 30 o 2 Reference level 1 TA  48.02   mVA 2  48.02     1 2 2  VA  VA 2 V Divide  2    1 :  4     4  A  2   VA  2VB     3  2 VB VB  VB 

 iii 

And

TB  12.005  

a  - g Sin  -9.8  Sin 30 o  -4.9

t  2  and

VB  VA  a t

 VB  VA  4.9  2    VB  VA  9.8     4  , then substitute  3  in  4   VB  2VB  9.8   VB  9.8 m / s then  VA  19.6 m / s

 iv  from 1 : m 

12.005

0.5  9.8 

2



1 kg 4

-------------------------------------------------------------------------------------------------------

Dynamic – 3rd secondary

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Chapter Six – work - Energy