Ch 14 Kinetics of Particles Work and Energy

Work and Energy 2

Work is defined as

U12   1

Ft

F .dr [ N .m  J ],  Fext .

dU12  F.dr  [Fxdx  Fy dy  Fz dz ]



1

1

Fext.  ma

U12   2

U12   1

r  dr

y

x O Horizental plane

U12   F.dr   ( Fx dx  Fy d y  Fz dz ) Where, 2

s2

s1

The total work along the path is 2

r F

dU  F.dr  ( F cos( ))ds  Ft ds, 2

ds dr

2

2

2

F .dr   Ft ds   m { a . dr }   mat ds 1  Fext . 1 1 1 (at et anen ) dset 2 1 1 m at ds   mvdv  [ mv22  mv12 ] 2 2 1 at ds vdv

U12  T2  T  T , [ J , lb. ft ]

T2

T1

T  (1/ 2)mv2 : 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

s

F

Example:

The force F is constant 2



2

U12   F.dr   F cos( )dr  F cos( ) L 1

dr

s1

s2

1

F cos( ) L  T2  T1 x

Work done by Spring Force

The work done by a spring force is 2

P  kx

2

1 1 U12   F . dr   kx.dx  kx22  kx12 2 2 1 kxi .dxi 1 Vs ,2

x1

Vs ,1

This work is the area under the force deflection y2 curve. It is stored in the spring as a potential energy Vs (elastic energy),. Vs ( x)  12 kx2 The

k

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.

P( x )

x2

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 y 2

U12   F.dr  1

U12  U12 

x2

 Fx .dx 

x1

x2

 Fx dx 

x1

y2



fdy 

y2

y1

 mgdy

y1

y2

y2

x2

y2

y2

x1

y1

y1

x1

y1

y1

Fy  mg  f

 max dx   mgdy   may dy   mvx dvx   mgdy   mvy dvy

y2

1 1 m (vx22  vx21 )  m(v2y 2  v2y1 )  mg ( y2  y1 ) 1 1 ;U12  m (v x22 v y22 )  m (v x22 v y22 )  mg ( y 2  y 1 ) 2 2 Work done 2 2 Work done Work done 1 1 m(v22 )  m(v12 )  (mgy2  mgy1 ) 2 2 T1

change of kinetic energy

1 m(v2 ) 2

Vg ,2

Fx y1

by gravity

to change vy

U12  T2  T1  (Vg ,2  Vg ,1 ) T

Vg ,1

g

F

x2

T2

T



Fy .dy 

y1

to change vx

U12 

y2

Vg

T1  Vg ,1  U12  T2  Vg ,2 E1

mg

x

E1  U12  E2

E2

change of potential energy due to gravity

: is called kinetic energy. E  T  Vg

Vg  mgy

: Work of the external force

E: Total Energy

b) If the force F is directed in the y-direction and equals the weight , then U12  Vg 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 Vg  Vg ,2  Vg ,1 not change, since it depends on the elevations only

y

Fy  mg  f

F y2

Fx y1

mg

x

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. y v1 e) If , U12  0 then T  V  0 , this the case of free fall y1 mg 1 1 T1  Vg1  T2  Vg 2 , T1  T2  Vg 2  Vg1, m v12  m v22  m g ( y2  y1 ), v 2 2 y2 2 x 2 2 v1  v2  2gy compare with v12  v22  2as from ch. 2.

y

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

Fy  mg  f

U12  T2 T1  ((V g ,2 V s ,2 )  (V g ,1 V s ,1 )); T1  V1  U12  T2  V2 T

V

V

E1

E2

V

E1  U12  E2 E  V  T dU dr J P   F .  F . v , [  W  Watt ], Power dt dt s hp  550 ft.lb / s  33,000 ft.lb / minute Pout e  , eoverall  em .ee Efficiency: m Pin

y2

F l 2 Fx y1

l1

hp  746W  .746 kW

x

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.

0.8 m

y Datum

A

h

E1  U12  E2

B 1 1 (V s V g )1  mV12  U12  (V s V g )2  mV 22 2 2 1 1 1 k (l1 )2  mg (0)  0  0  k (l2 )2  mg (h )  mV 22 2 2 2

l1  0.8  0.4  0.4m , l2  [0.82  0.62 ]0.5  0.4]  0.6 m 1 1 1 (200)(0.4)2  3(9.81)(0)  0  0  (200)(0.6)2  3(9.81)(0.6)  (3)V 22 2 2 2 V 2  1.53 m / s

0.6 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. v1  0, v2  ? 1 k (x)2  0  2

1 2 1 mv1  ( Vg  Vs )1  U12  mv22  (Vg  Vs )2 (Vs )1  (Vg  Vs )2 2 2 (Vs )2 1 = (500)( h2 0.05)2 2 Assume the plunger leaves the spring, h 50 mm

 mgh

1 1 (500)(.10)2  (500)(0.05)2  2(9.81)( h), h  0.0956 m  95.6 mm 2 2

Since h > 50mm , your assumption is correct.

Example 13: The system is released from rest from the shown position o  60 , v  0   180 , vsphere  ??

6 kg

0.3 m

 E1  U12   E2  [V1  T1 ]  U12   [V 2 T2 ]  [(Vs V g )1]   [(Vs V g )2 T2 ]  (V sphere Vcyl )1 

y

0.3 m

Datum 0.3 m

 0.3 m

0.3 m

6 g (0.3)sin(30)  4 g (2)(0.3)sin(30)  -2.943 J



(Vspher  Vcyl )2  6 g (0.3)  4 g (2)(0.3)  -5.886 J

 

1 1 2 2 1 1mcyl T  m v  v 2 2 2 spher spher cyl T2  2 mspherv spher 2 mcyl v cyl 2 2 Top dead center

0.3 m

4 kg

0.3 m

Bottom dead center

1 2 -2.943+0=-5.886+ (6)vsphere , vsphere  0.99 m / s 1 2 2 -2.943=-5.886+ (6)v sphere , v sphere  0.99 m / s 2

v sphere  0.99 m / s

2

1

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. [ (V g )1  T1 ]  U12  [(V g )2 T2 ]

U12  F .L  F (L1  L2 )  F ([ 8002  2002 /1000]  [ 2002  2002 /1000]) U12  0.54178F .

1 T2  (0.6)(v 22 )  0.3(16)  4.8 J 2 V 2  mgh  0.6(9.81)(0.4)  2.3544J 0.54178F .  4.8  2.3544,

F  13.21N

C 12 13 Solved problems

,

Problem 13.11 The 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, v = 2.857 m/s

v2 = 8.1615, WA = 7(9.81) = 68.67 N

C 12 13 Solved problems Problem 13.27 A 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

9 2 U12   (1.2  9 x)dx  0.5 x  0.7 x  x 0 2 9 2 2 v  4 g [0.7x  x ] 2 x

C 12 13 Solved problems

9 2 2 v  4 g [0.7x  x ]

2 dv 9 2 v max   0  0.7x  x  x  0.077778ft dx 2 9 2 v max  4 g [0.7(0.077778)  (0.077778) 2 ]  3.5 2 v max  1.872m / s

C 12 13 Solved problems Problem 13-29 A 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 condition Position 2, spring deflected 5 inches Position 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 unstretched 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 x B  x c  d x B  x c x c  (x c  x A )  l

2x c  x A

Rename x A  x A , x B  x B , x c  x c U 1 2 W x A  0.5k A x A2  0.5k B x B2  T 2  T1  0.5m Av 2  98.1(0.05)  0.5(2000)(0.05)2  0.5(2000)(0.025)2  0.5(10)v 2 ,x B v  0.597m / s Another method: E1  E 2  E  0, Vs  T  V g  0 2  0)  (0.5k x 2  0) W x  0.5m (v 2  0)  0 (0.5k A x A B B A A A 2

0.5(2000)(0.05)2  0.5(2000)(0.025)2  98.1(0.05)  0.5(10)v 2 v  0.597m / s

Datum xc xA

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.

v2 mg  T  m l

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.