F

1s

2s

stF

: is the kinetic energy of the particle

Thus, the work of all external forces (including the gravity, spring forces) is equal to the particle’s kinetic energy change

Work and Energy

Work is defined as .

2

1 21

.ext

U d

F

F r

dsdr

O x

y

rdr r

Horizental plane

1 2 . [ ]x y zdU d F dx F dy F dz F r

. ( cos( )) ,tdU d F ds Fds F r

[ . ],N m J

The total work along the path is2 2

1 21 1

. ( )x x y y z zU d F d F d F d F r

.

2 2

1 21 1

.ext

tU d Fds

F

F r.ext mF a 2 2

1 1( )

{ . }tt t n n

tdsea e a e

m d ma ds

a r

2 1

2 22 2

1 2 2 11 1

1 1[ ]2 2

t

t

a ds vdvT T

U m a ds mvdv mv mv

1 2 2 , [ , . ]U T T T J lb ft 2(1/ 2)T mv

Where,

This work is the area under the force deflection curve. It is stored in the spring as a potential

energy

Example:The force F is constant

F

dr

2s1s2 2

1 21 1

2 1

. cos( ) cos( )

cos( )

U d F dr F L

F L T T

F r

x

P kxk

( )P x

1x 2x

Work done by Spring Force

,2 ,1

2 22 2

1 2 2 1.1 1

1 1. .

2 2s s

kx dxV V

U d kx dx kx kx i i

F r

sV

21( )

2sV x kx

The work done by a spring force is

(elastic energy),.

The amount of the work done on (by) the spring depends on the initial and final points not on the

followed path, All forces whose work do not depend on the path of motion are called conservative forces.

2y

Kinetic and Potential energy of Gravity:a) If a particle is elevated from y1 to y2 ; the work done by the force to move the particle from point y1 to point y2 is

2 2 2 2 2

1 1 1 1 1

2

1 21

. . .x y x y y

x y xx y x y y

U d F dx F dy F dx fdy mgdy F r

2 2 2 2 2 2

1 1 1 1 1 1

1 2

x y y x y y

x y x x y yx y y x y y

U ma dx mgdy ma dy mv dv mgdy mv dv

mg1y

2y

x

y

yF mg f

xF

F g

2 1 2 1

x y

2 2 2 21 2 2 1

Work done Work done Work done by gravityto change v to change v

1 1( ) ( ) ( )

2 2x x y yU m v v m v v mg y y 2 2 2 2

2 2 2 21 2 2 1

1 1; ( ) ( ) ( )

2 2x y x yU m v v m v v mg y y

,2 ,1

2 1

2 21 2 2 1 2 1

change of potential change of kinetic energy due to gravity energy

1 1( ) ( ) ( )

2 2g gV V

T T

U m v m v mgy mgy

1 2 2 1 ,2 ,1( )g g

T Vg

U T T V V

1 2

1 ,1 1 2 2 ,2g g

E E

T V U T V 1 1 2 2E U E

21( )

2T m v : is called kinetic energy. gV mgy : Work of the external force

gE T V E: Total Energy

b) If the force F is directed in the y-direction and equals the weight , then 1 2 gU V

mg1y

2y

x

y

yF mg f

xF

F

c) If we changed path of motion of the particle, but still it starts at point 1 and reaches point 2, the work done by the gravity will not change, since it depends on the elevations only ,2 ,1g gVg V V

d) Forces whose work depends on the initial and final points “not the followed path” are call conservative forces, thus gravity force is conservative force. On the other hand

forces whose work depends on the path followed by the particle are called non conservative forces, such as the friction forces. The work of friction force dissipates as

energy lost.

e) If , then , this the case of free fall1 2 0U 0T V mg

2y

1y

x

y1v

2v1 1 2 2 1 2 2 11

, ,2g g g gT V T V T T V V m 2

11

2v m 2

2v m 2 1( ),g y y

2 21 2 2v v g y compare with from ch. 2. 2 2

1 2 2v v a s

f) if the particle were attached with spring as shown, then

2y

x

y

yF mg f

xFF

1l

2l

1y

1 2 2 1 ,2 ,2 ,1 ,1(( ) ( ));g s g s

T V V

V

U T T V V V V

1 2

1 1 1 2 2 2

E E

T V U T V

1 1 2 2E U E E V T

. . , [ ],dU dr J

P F F v W Wattdt dt s

746 .746hp W kW

550 . / 33,000 . / minutehp ft lb s ft lb

Power

Efficiency: , .outm overall m e

in

Pe e e e

P

1 1 2E U 2E

21 1

1( )

2s gV V mV 1 2U 22 2

1( )

2s gV V mV

Example 12: The spring has an unstretched length of 0.4 m and a stiffness of 200N/m. The

3-kg slider and attached spring are released from rest at A and moves in the vertical plane.

Calculate the velocity v of the slider as it reaches B in the absence of friction.

Datum

y

h

A

B

0.8m

0.6m

2 2 22

2

1 1 1(200)(0.4) 3(9.81)(0) 0 0 (200)(0.6) 3(9.81)( 0.6) (3)

2 2 21.53 /

V

V m s

2 2 21 2 2

1 1 1( ) (0) 0 0 ( ) ( )

2 2 2k l mg k l mg h mV

2 2 0.51 20.8 0.4 0.4 , [0.8 0.6 ] 0.4] 0.6l m l m

Example 13:The 2-kg plunger is released from rest in the position shown where the spring of stiffness

k=500 N/m has been compressed to one half its uncompressed length of 200 mm. Calculate

the maximum height h above the starting position reached by the plunger.

1 20, ?v v 21

1

2mv ( gV 1 1 2)sV U 2

21

2mv 2( )g sV V 1 2( ) ( )s g sV V V

2

2

22

1= (500)( 0.05)

2Assume the plunger leaves the spring, 50

1( ) 0 ( )

2 s

h

h mm

k x V mgh

2 21 1(500)(.10) (500)(0.05) 2(9.81)( ), 0.0956 95.6

2 2h h m mm

Since h > 50mm , your assumption is correct.

Example 13: The system is released from rest from the shown position

60 , 0180 , ??

o

sphere

vv

Datum

0.3 m

0.3 m

6 kg

4 kg

0.3 m

0.3 m

0.3 m

0.3 m

0.3 m

y

Example 14:The 0.6-kg collar slides on the curved rod in the vertical plane with negligible friction under the action of constant force F in the cord guided

by the small pulleys at D. If the collar is released from rest at A, determine the

force F which will result in the collar striking the stop at B with a velocity of

4m/s.

1[ ( )gV 1T 1 2 2 2] [( ) ]gU V T

2 2 2 21 2 1 2. ( ) ([ 800 200 /1000] [ 200 200 /1000])U F L F L L F

0.54178 . 4.8 2.3544,F

1 2 0.54178 .U F

22 2

1(0.6)( ) 0.3(16) 4.8

2T v J

2 0.6(9.81)(0.4) 2.3544V mgh J

13.21F N

C 12 13 Solved problems

,

Problem 13.11The 7-kg block A is released from rest in the position shown. Neglecting the effect of friction and the masses of the pulleys, determine the velocity of the block after it has moved 0.6 m up the incline.

d2=(1.2)2+(0.6)2−2(1.2) (0.6) cos 15°

d2= 0.63958 m2

UC=FC (Distance pulley C lowered)=(140 )(1/ 2)(1.2 -0.63958) =39.229 N m

UA= −68.67 (sin15°) (0.6) = −10.6639 N m⋅

U=T2−T1=UC−UA

1/2 mAv 2=UC−UA

(1 /2) (7) v2 =(39.229 -10.6639) N m, v2 = 8.1615, v = 2.857 m/s WA = 7(9.81) = 68.67 N

C 12 13 Solved problems

Problem 13.27A 0.7-lb block rests on top of a 0.5-lb block supported by but not attached to a spring of constant 9 lb/ft. The upper block is suddenly removed. Determine (a) the maximum velocity reached by the 0.5-lb block, (b) the maximum height reached by the 0.5-lb block.

At the initial position (1), the force in the spring equals the weight of both blocks, i.e., 1.2 lb.

Thus at a distance x, the force in the spring is,

Fs=1.2−kx Fs=1.2−9x

Max velocity of the 0.5 lb block occurs while the spring is still in contact with the block.

T 1=0, T2=(1/2)mv2=(1/2)(0.5/g) v2

2

021 2

97.05.0)92.1( xxxdxxU

x

2 294 [0.7 ]

2v g x x

C 12 13 Solved problems

2 294 [0.7 ]

2v g x x

2max

90 0.7 0.077778

2

dvv x x x ft

dx

2 2max

94 [0.7(0.077778) (0.077778) ] 3.5

2v g

max 1.872 /v m s

C 12 13 Solved problems

Problem 13-29A 7.5-lb collar is released from rest in the position shown, slides down the inclined rod, and compresses the spring. The direction of motion is reversed and the collar slides up the rod. Knowing that the maximum deflection of the spring is 5 in., determine (a) the coefficient of kinetic friction between the collar and the rod,(b) the maximum speed of the collar.

Position 1, initial conditionPosition 2, spring deflected 5 inchesPosition 3, initial contact of spring with collar

C 12 13 Solved problems

C 12 13 Solved problems

Problem 13. 30:A 10-kg block is attached to spring A and connected to spring B by a cord and pulley. The block is held in the position shown with both springs un-stretched when the support is removed and the block is released with no initial velocity. Knowing that the constant of each spring is 2 kN/m, determine (a) the velocity of the block after it has moved down 50 mm, (b) the maximum velocity achieved by the block.

(a) W = Weight of the block =10(9.81)=98.1 N

B cx x d B cx x ( )c c Ax x x l 2 c Ax x

2 21 2 2 10.5 0.5A A B BAU Wx k x k x T T 2

2 2 2

0.5

98.1(0.05) 0.5(2000)(0.05) 0.5(2000)(0.025) 0.5(10) ,

0.597 /

Am v

v

v m s

2 2 22(0.5 0) (0.5 0) 0.5 ( 0) 0A B B A A AAk x k x W x m v

1 2 0, 0gE E E Vs T V

2 2 20.5(2000)(0.05) 0.5(2000)(0.025) 98.1(0.05) 0.5(10)

0.597 /

v

v m s

Rename , ,A A B B c cx x x x x x

Another method:

Ax

Bx

cx

Datum

C 12 13 Solved problems

(b) Let x = Distance moved down by the 10 kg block

C 12 13 Solved problems

Problem 13, 41:The sphere at A is given a downward velocity v0 and swings in a vertical circle of radius l and center O. Determine the smallest velocity v0 for which the sphere will reach point B as it swings about point O (a) if AO is a rope, (b) if AO is a slender rod of negligible mass.

l

vmTmg

2

C 12 13 Solved problems

Problem 13, 44: A section of track for a roller coaster consists of two circular arcs AB and CD joined by a straight portion BC. The radius of AB is 27 m and the radius of CD is 72 m. The car and its occupants, of total mass 250 kg, reach point A with practically no velocity and then drop freely along the track. Determine the normal force exerted by the track on the car as the car reaches point B. Ignore air resistance and rolling resistance.

C 12 13 Solved problems

A 6 kg collar is attached to a spring anchored at point C and can slide on a frictionless rod forming an angle of 30 degrees with the vertical. The spring is of constant k and is unstretched when the collar is at A. Knowing that the collar is released from rest at A, calculate the speed of the collar at point B for values of k from 20N/m to 400N/m.