8/4/2019 Mech103 Tut Ch15

1/22

Problem 15.9 As the 10,000-kg airplane takes off,the tangential component of force exerted on it by itsengines is constant and equal to 60 kN. Neglectingother forces acting on the airplane, use the principle ofwork and energy to determine its velocity when it has

rolled 500 m.

Solution:

U12 = (60 kN)(500 m) =1

2(10,000 kg)v2 v = 77.5 m/s

Principle of Work and Energy

8/4/2019 Mech103 Tut Ch15

2/22

Problem 15.10 Determine (a) the maximum powerand (b) the average power transferred to the airplane inProblem 15.9 during its 500-m roll.

Solution:

(a) The maximum power occurs when the plane reaches its maximum

speed

Pmax = (60 kN)(77.5 m/s) = 4.65 MW

(b) The average power is the total work divided by the total time.

Using F = ma

60 kN = (10,000 kg)a, a = 6 m/s2, v = (6 m/s2)t, s = (3 m/s2)t2

500 m = (3 m/s2)t2 t= 12.9 s

Pave =(60 kN)(500 m)

12.9 s= 2.32 MW

8/4/2019 Mech103 Tut Ch15

3/22

Problem 15.11 When aerodynamic drag is accountedfor, the tangential component of force exerted on theairplane in Problem 15.9 is given as a function of thedistance s (in meters) traveled by the airplane by

Ft = 60e0.0024s kN.

Use the principle of work and energy to determine itsvelocity when it has rolled 500 m.

Solution:

U=

500 m0

(60 kN)e0.0024sds = 1.747 107 N-m

1.747 107 N-m =1

2(10,000 kg)v2

v = 59.1 m/s

8/4/2019 Mech103 Tut Ch15

4/22

Problem 15.14 The force exerted on a car by aprototype crash barrier as the barrier crushes isF = (4400+ 50,000s) N, where s is the distance inmeters from the initial contact. Suppose that you wantto design the barrier so that it can stop a 2400-kg car

traveling at 100 km/h. What is the necessary effectivelength of the barrier? That is, what is the distancerequired for the barrier to bring the car to a stop?

s

Solution:

sf0

F ds =1

2mv2f

1

2mv20

vf = 0

v0 = 100km

h= 27.78

m

s

sf0

(4400+ 50,000 s) ds = 1

2mv20

(4400 s + 25,000 s2)

sf0=

1

2(2400)(27.78)2

Solving sf = 6.00 m

8/4/2019 Mech103 Tut Ch15

5/22

Problem 15.15 The duration of the impact of the carwith the barrier described in Problem 15.14 is 0.33 s.What average power is transferred from the car duringthe impact?

Solution:From the solution to Problem (15.14)

Work= 1

2mv20 =

1

2(2400)(27.78)2

U12 = Work= 9.26 105 N-m

P =U12

t2 t1=9.26 105

0.33 s

P = 2.81 106 watts

P = 2.81 MW

8/4/2019 Mech103 Tut Ch15

6/22

Problem 15.28 The masses of the three blocks aremA = 40 kg, mB = 16 kg, and mC = 12 kg. Neglect themass of the bar holding C in place. Friction is negligible.By applying the principle of work and energy to A and Bindividually, determine the magnitude of their velocity

when they have moved 500 mm.

A

B

C

45

Solution: Denote b = 0.5 m. Since the pulley is one-to-one,

denote |vA| = |vB | = v. The principle of work and energy for weight

A is

b0

(mAg sin T)d s =1

2mAv

2,

and for weight B

b0

(TmBg sin )ds =1

2mBv

2.

Add the two equations:

(mA mB )gb sin =12(mA +mB )v

2.

T

T

NBC

NAB

NAB

NA

mBg

mAg

8/4/2019 Mech103 Tut Ch15

7/22

Problem 15.29 Solve Problem 15.28 by applying theprinciple of work and energy to the system consisting ofA, B , the cable connecting them, and the pulley.

Solution: Choose a coordinate system with the origin at the pulley

axis and the positive x axis parallel to the inclined surface. Since the

pulley is one-to-one, xA = xB . Differentiate to obtain vA = vB .

Denote b = 0.5 m. From the principle of work and energy the work

done by the external forces on the complete system is equal to the

gain in kinetic energy,

xA0

mAg sin d s +

xB0

mB g sin d s =1

2mAv

2A +

1

2mB v

2B ,

from which

(mB mA)gb sin =12

(mA +mB )v2A

and |vA| = |vB | =

(mA mB )

(mA +mB )2gb sin = 1.72 m/s.

NBC

NA

mBg

mAg

8/4/2019 Mech103 Tut Ch15

8/22

Problem 15.30 In Problem 15.28, determine the mag-nitude of the velocity ofA and B when they have moved500 mm if the coefficient of kinetic friction between allsurfaces is k = 0.1.

Strategy:The simplest approach is to apply the prin-ciple of work and energy to A and B individually. If

you treat them as a single system, you must account forthe work done by internal friction forces.

Solution: Since the pulley is one-to-one, |vA| = |vB | = v. Denoteb = 0.5 m. The principle of work and energy applied to weight A is

b

0

(mAg sin T kNA kNAB ) ds =1

2

mAv2,

and for weight B

b0

(TmBg sin kNBC kNAB ) ds =1

2mBv

2.

Add the two equations,

(mAm

B)gb sin

kb(N

A+ 2N

AB+N

BC) = 1

2(m

A+m

B)v2.

The normal forces are

NBC = mCg cos ,

NAB = (mB +mC)g cos ,

mAg

mBg

NA

NBC

NAB

NAB kNBC

kNAB

kNAB

kNA

T

T

8/4/2019 Mech103 Tut Ch15

9/22

Problem 15.50 Suppose that you want to design abumper that will bring a 50-lb. package moving at 10 ft/sto rest in 6 in from the point of contact with bumper.If friction is negligible, what is the necessary springconstant k? k

10 ft/s

Solution: From the principle of work and energy, the work done

on the spring must equal the change in kinetic energy of the package

within the distance 6 in.

1

2kS2 =

1

2

W

g

v2

from which

k =

W

g

vS

2=

50

32.17

10

0.5

2= 621.7 lb/ft

Work Done by particular Forces

8/4/2019 Mech103 Tut Ch15

10/22

Problem 15.51 In Problem 15.50, what spring con-stant is necessary if the coefficient of kinetic frictionbetween the package and the floor is k = 0.3 and thepackage contacts the bumper moving at 10 ft/s?

Solution: The work done on the spring over the stopping dis-

tance is

US =

S0

F ds =

S0

ks ds =1

2kS2.

The work done by friction over the stopping distance is

Uf =

S0

F ds =

S0

k W ds = k W S.

From the principle of work and energy the work done must equal the

kinetic energy of the package:

1

2kS2 +k W S=

1

2

W

g

v2,

from which, for S= 12

ft,

k =

W

g

(v2 2 gk S)

S2= 561.7 lb/ft

8/4/2019 Mech103 Tut Ch15

11/22

Problem 15.63 The 4-kg collar is released from restat position 1. Neglect friction. If the spring constant isk = 6 kN/m and the spring is unstretched in position 2,what is the velocity of the collar when it has fallen toposition 2?

k

2

1

200 mm

250 mm

Solution: Denote d= 200 mm, h = 250 mm. The stretch of thespring in position 1 is S1 =

h2 + d2 d= 0.120 m and at 2 S2 = 0.

The work done by the spring on the collar is

Uspring = 00.12

(ks)ds = 12ks20

0.120= 43.31 N-m.

The work done by gravity is

Ugravity =h

0

(mg)ds = mgh = 9.81 N-m.

8/4/2019 Mech103 Tut Ch15

12/22

Problem 15.64 The 4-kg collar is released from restin position 1 on the smooth bar. The spring constant isk = 4 kN/m. The tension in the spring in position 2 is500 N. What is the velocity of the collar when it hasfallen to position 2?

Solution: Denote d= 200 mm, h = 250 mm. The stretch of thespring at position 2 is

S2 =T

k= 500

4000= 0.125 m.

The unstretched length of the spring is L = d S2 = 0.2 0.125 =0.075 m. The stretch of the spring at position 1 is S1 =

h2 + d2

L

=0.245 m. The work done by the spring is

Uspring =S2S1

(ks)ds = 12k(S21 S22 ) = 88.95 N-m.

The work done by gravity is Ugravity = mgh = 9.81 N-m. From theprinciple of work and energy is Uspring +Ugravity = 12mv2, from which

v =

2(Uspring +Ugravity)m

= 7.03 m/s

8/4/2019 Mech103 Tut Ch15

13/22

Problem 15.65 The 4-kg collar starts from rest inposition 1 on the smooth bar. Its velocity when it hasfallen to position 2 is 4 m/s. The spring is unstretchedwhen the collar is in position 2. What is the springconstant k?

Solution: The kinetic energy at position 2 is 12mv2 = 32 N-m.

From the solution to Problem 15.63, the stretch of the spring in posi-

tion 1 is S1 =h2 + d2 d= 0.120 m. The potential of the spring is

Uspring =0S1

(ks)ds = 12kS21 .

The work done by gravity is Ugravity = mgh = 9.81 N-m. From theprinciple of work of work and energy, Uspring

+Ugravity

=12mv2. Sub-

stitute and solve:

k = 2

12mv2 Ugravity

S21

= 3082 N/m

8/4/2019 Mech103 Tut Ch15

14/22

Problem 15.71 The 10-kg collar starts from rest atposition 1. The tension in the string is 200 N, and they-axis points upward. If friction is negligible, what is themagnitude of the velocity of the collar when it reachesposition 2?

y

200 N

x

(4, 4, 2) m

(6, 2, 1) m

(1, 1, 0) m1

2

z

Solution: The constant force moves a distance

s =

(6 1)2 + (2 1)2 + (1 0)2

(6 4)2 + (2 4)2 + (1 2)2 = 2.2 m.

The work done by the constant force is

UF =

s0

F ds = F s = 439.2 N-m.

The work done by gravity is

Ugravity =

h0

(mg)ds = mgh = (10)(9.81)(3)

= 294.3 N-m.

8/4/2019 Mech103 Tut Ch15

15/22

Problem 15.76 A satellite in a circular orbit of

radius r around the earth has velocity v =

gR2E /r ,

where RE = 6370 km is the radius of the earth.Suppose you are designing a rocket to transfer a900-kg communication satellite from a parking orbit

with 6700-km radius to a geosynchronous orbit with42,222-km radius. How much work must the rocket doon the satellite?

Solution: Denote the work to be done by the rocket by Urocket .

Denote Rpark = 6700 km, Rgeo = 42222 km. The work done by the

satellites weight as it moves from the parking orbit to the geosyn-

chronous orbit is

Utransfer =Rgeo

Rpark

F ds =Rgeo

Rpark

mg R

2E

s2ds

=

mg

R2E

s

RgeoRpark

= mgR2E

1

Rgeo

1

Rpark

,

Utransfer = 4.5 109 N-m. From the principle of work and energy:

Utransfer +Urocket =

12

mv2

geo

12

mv2

park

.

from which

Urocket =

1

2mv2

geo

1

2mv2

park

Utransfer .

8/4/2019 Mech103 Tut Ch15

16/22

Problem 15.94 The mass m = 1 kg, the spring con-stant k = 200 N/m, and the unstretched length of thespring is 0.1 m. When the system is released from restin the position shown, the spring contracts, pulling themass to the right. Use conservation of energy to deter-mine the magnitude of the velocity of the mass whenthe string and spring are parallel.

k

0.3 m

0.15 m

0.25 m

Solution: The stretch of the spring in position 1 is

S1 =(0.15)2 + (0.25)2 0.1 = 0.192 m.

The stretch in position 2 is

S2 =(0.3+ 0.15)2 + (0.25)2 0.3 0.1 = 0.115 m.

The angle = arctan(0.25/0.45) = 29.1

. Applying conservation of

energy,

12mv21 +

12kS21 mg(0.3) =

12mv22 +

12kS22 mg(0.3cos ):

0.3 m

0.15 m

0 25 m

2

1

Datum

Conservation of Energy

8/4/2019 Mech103 Tut Ch15

17/22

Problem 15.95 In problem 15.94, what is the tensionin the string when the string and spring are parallel?

Solution: The free body diagram of the mass is: Newtons second

law in the direction normal to the path is

T kS2 mg cos = man:

T (200)(0.115) (1)(9.81) cos29.1= (1)(v22/0.3).

We obtain, T = 44.7 N.

8/4/2019 Mech103 Tut Ch15

18/22

Problem 15.96 The force exerted on an object bya nonlinear spring is F = [k(r r0)+ q(r r0)

3]er ,where k and q are constants and r0 is the unstretchedlength of the string. Determine the potential energy ofthe spring in terms of its stretch S= r r0.

k

r

Solution: Note that dS= dr . The work done in stretching the

spring is

V = F dr +C =

F (dr er + rd e )+C

=

[k(r r0)+ q(r r0)

2] dr + C,

V =

[kS+ qS3] dS+ C.

Integrate:

V =k

2S2 +

q

4S4 ,

where C = 0, since F = 0 at S= 0.

8/4/2019 Mech103 Tut Ch15

19/22

Problem 15.99 The string exerts a force of constantmagnitude T on the object. Use polar coordinates toshow that the potential energy associated with this forceis V = T r .

T

r

Solution:

dV = F dr

V =

r

DATUM

Ter dr er

V = T r

rDATUM

V = T r T rDATUM

Let rDATUM = 0

V = T r

T

er

8/4/2019 Mech103 Tut Ch15

20/22

Problem 15.100 The system is at rest in the positionshown, with the 12-lb collar A resting on the spring(k = 20 lb/ft), when a constant 30-lb force is applied tothe cable. What is the velocity of the collar when it hasrisen 1 ft? (See Problem 15.99.)

30 lb

3 ft

2 ft

A

k

Solution: Choose the rest position as the datum. At rest, the com-

pression of the spring is

S1 =Wk

= 0.6 ft.

When the collar rises 1 ft the stretch is S2 = S1 + 1 = 0.4 ft Whenthe collar rises 1 ft the constant force on the cable has acted through

a distance

s = 32 + 22 (3 1)2 + 22 = 0.7771 ft.The work done on the system is Us = 12 k(S21 S22 )mg(1)+ Fs.From the conservation of energy Us = 12mv2 from which

v =

k

m(S21 S22 ) 2g +

2Fs

m= 8.45 ft/s.

8/4/2019 Mech103 Tut Ch15

21/22

Problem 15.155 The y-axis is vertical and the curvedbar is smooth. If the magnitude of the velocity of the4-lb slider is 6 ft/s at position 1, what is the magnitudeof its velocity when it reaches position 2?

y

x

1

2

2 ft

4 ft

Solution: Choose the datum at position 2. At position 2, the energycondition is

1

2

W

g

v21 +Wh =

1

2

W

g

v22,

where h = 2, from which

v2 = v21 + 2 gh = 6

2 + 2 g(2) = 12.83 ft/s

8/4/2019 Mech103 Tut Ch15

22/22

Problem 15.156 In Problem 15.155, determine themagnitude of the velocity of the slider when it reachesposition 2 if it is subjected to the additional force F =3xi 2j (lb) during its motion.

Solution:

U=

F dr =

02

(2) dy +

40

3x d x

= [2y]02 +

3

2x24

0

= 4 + 24 = 28 ft-lb.

From the solution to Problem 15.155, the energy condition at posi-

tion 2 is

1

2

W

g

v21 + W h+ U=

1

2

W

g

v22 ,

from which

v2 =

v21 + 2gh +

2g(28)

W=

62 + 2g(2) +

2g(28)

4

= 24.8 ft/s