# Chapter 17: Work and Energy¶

## Example 17.17-4, Page No 400¶

In :
# Initilization of variables
s1=2 # compression of the spring- initial
s2=5 # compression of the spring- final

import math
def integrand(s, a, b):
return 20*s
a=1
b=1

# Results
print'The word done in compressing the spring is',round(U),"in-lb"

The word done in compressing the spring is 210.0 in-lb


## Example 17.17-6, Page No 400¶

In :
import math

#Initilization of variables
m=5 #kg
d=6 #m
# as theta1=30 degrees & theta2=10 degrees,
sintheta1=2**-1
sintheta2=0.1736
costheta1=(3**0.5)*2**-1
costheta2=0.9848
u=0.2 #coefficient of friction
g=9.8 #m/s**2
F=70 #N

#Calculations
#Using free body diagram
Na=(m*g*costheta1)-(F*sintheta2) #N
#work done by each force
W=[F*costheta2,-m*g*sintheta1,0,-u*Na*d] #N.m
#Total Work Done
W_tot=W+W+W+W #N.m
#Using resultant
R=F*costheta2-(u*Na)-(m*g*sintheta1) #N
W_d=R*d #N.m (Work Done)

#Result
print'The work done is',round(W_d),"N.m"

The work done is 230.0 N.m


## Example 17.17-7, Page No 401¶

In :
import math

#Initilization of variables
m=20 #kg
d=1.5 #m
# as theta=30 degrees,
sintheta=2**-1
u=0.25 #coefficient of friction
g=9.8 #m/s**2
F=130 #N

#Calculations
W=F*d-(m*g*sintheta*d) #N.m

#Result
print'The work done is',round(W),"N.m"

The work done is 48.0 N.m


## Example 17.17-9, Page No 402¶

In :
import math

#Initilization of variables
d=6*12**-1 #ft
l=8*12**-1 #ft
l_c=3.2 #in
y=1.82 #in**2

#Calculations
V=1*4**-1*pi*d**2*l #ft**3
#One horizontal inch
h_i=V/l_c #ft**3
#One vertical inch
v_i=100*144 #lb/ft**2
#Then 1.82 in**2 represents
x=y*v_i*h_i #ft-lb

#Result
print'The work capacity is',round(x),"ft-lb"

# The ans in the textbook is incorrect.

The work capacity is 1072.0 ft-lb


## Example 17.17-10, Page No 402¶

In :
import math

#Initilization of variables
speed=90000 #m/h
P=100*1000 #N

#Calculations
Power=P*((speed)/3600) #J/s

#Result
print'The power developed is',round(Power*10**-6,1),"MJ/s"
# Note the unit used.

The power developed is 2.5 MJ/s


## Example 17.17-11, Page No 403¶

In :
import math

#Initilization of variables
d=0.6 #m
T_t=800 #N
T_s=180 #N
w=200 #rpm

#Calculations
#Torque
M=(T_t-T_s)*r #N.m
#Power
Power=M*(w_new) #W

#Result
print'The power transmitted is',round(Power*10**-3,1),"kW"

# The answer in the book is incorrect

The power transmitted is 3.9 kW


## Example 17.17-12, Page No 403¶

In :
import math

#Initilization of variables
P=25.6 #lb
w=600 #rpm
a=36 #in
b=12 #in

#Calculations
M=P*(((b*2**-1)+a)/12)  #lb-ft
Hp=(M*w_new)/550 #hp

#Result
print'The power being transmitted is',round(Hp,1),"hp"

The power being transmitted is 10.2 hp


## Example 17.17-13, Page No 403¶

In :
import math

#Initilization of variables
Pout=3.8 #bhp
Pin=4.1 #ihp

#Calculations
Efficiency=round((Pout/Pin)*100) #Percent

#Result
print'The efficiency of the engine is',round(Efficiency),"%"

The efficiency of the engine is 93.0 %


## Example 17.17-15, Page No 404¶

In :
import math
def integrand(x, a, b):
return -(3/x)
a=1
b=1
g=32.2 # ft/s**2
deltaT=((U*32.2*2)/(4*16**-1))**0.5
# Results
print'The speed of the disc will be',round(deltaT,1),"ft/s"

The speed of the disc will be 23.1 ft/s


## Example 17.17-16, Page No 404¶

In :
import math

#Initilization of variables
l=2 #m
m=4 #kg
w_1=20 #rpm
w_2=50 #rpm
rev=10 #no of revolution

#Calculations
Io=(3**-1)*(m)*l**2 #kg.m**2
M=(0.5*Io*(w2**2-w1**2))/theta #N.m

#Result
print'The constant moment required is',round(M,3),"N.m"
# The ans waries in decimal places.

The constant moment required is 0.977 N.m


## Example 17.17-18, Page No 405¶

In :
import math

#Initilization of variables
W=1000 #lb
w_w=200 #lb weight of the individual wheel
d_w=2.5 #ft diameter of the wheel
v=22 #ft/s
t=2 #minutes

#Calculations
#T1=Initial Kinetic Energy and T2=Final Kinetic Energy
F=(-0.5*W*32.2**-1*v**2-4*0.5*w_w*32.2**-1*(v**2+0.5*v**2))/(10560) #lb

#Result
print'The rolling resistance is',round(F,2),"lb"

The rolling resistance is -1.57 lb


## Example 17.17-19, Page No 405¶

In :
import math

#Initilization of variables
W=100 #lb
lo=4 #ft
# as theta=45 degrees
costheta=(2**0.5)**-1
g=32.2 #ft/s**2
l=8/3 #ft

#Calculations
#Taking moment about point O and equating it to zero
#Summing forces in the t direction
Ot=(W*costheta)-((W/g)*lo*0.5*alpha) #lb
#Work Done
Work=W*(lo*0.5*costheta) #ft/lb
#Moment of inertia
Io=(3**-1)*(W/g)*(lo**2) #kg-ft**2
#Using the concept for work done=chane in K.E
#Summing forces along the bar
On=-(-((W/g)*lo*0.5*w**2)-(W*costheta)) #lb

#Result
print'The bearing reaction at O on the rod is',round(On),"lb"

The bearing reaction at O on the rod is 177.0 lb


## Example 17.17-21, Page No 406¶

In :
import math

#Initilization of variables
vo=9 #m/s
# as theta=30 degrees,
sintheta=2**-1
g=9.8 #m/s**2

#Calculations
x=((7*10**-1)*vo**2)*(g*sintheta)**-1 #m

#Result
print'The ball will roll',round(x,1),"m up the plane"

#The textbook wrongly mentions the unit of displacement as in

The ball will roll 11.6 m up the plane


## Example 17.17-22, Page No 406¶

In :
import math

#Initilization of variables
W=322 #lb
F=12 #lb
a=0 #lower limit (where the cyliner starts rolling)
b=pi/2 #Upper Limit (where the cyliner stops rolling)
d=3.2 #ft
g=32.2 #ft/s**2

#Calculations
d_U=2*dR*F #differential work done
#Integration Calculations
#As it is a simple integration we can resort to this
U=d_U*(b-a) #ft-lb
#Determination of K.E

#Result
print'The angular velocity of the cylinder is',round(w,2),"rad/s"

The angular velocity of the cylinder is 2.69 rad/s


## Example 17.17-24, Page No 407¶

In :
import math

#Initilization of variables
Wa=161 #lb
Wb=193.2 #lb
Wc=322 #lb
v1=5 #ft/s
lc=6 #in
k=6 #lb/ft
l=4 #ft
u=0.2 #coefficient of friction
g=32.2 #ft/s**2

#Calculations
Ib=(2**-1)*(Wb/g)*(2**-1)**2 #Moment of inertia
T1=(0.5*(Wc/g)*v1**2)+(0.5*Ib*w1**2)+(0.5*(Wa/g)*v1**2) #ft-lb
#Work Done on the system
#The textbook is ambigious on the calculations hence the result is dispalyed directly
U=26.4 #ft-lb
#Velocity Calculations
v=((T1+U)*9**-1)**0.5 #ft/s

#Result
print'The velocity of the block is',round(v,2),"ft/s"

The velocity of the block is 5.29 ft/s


## Example 17.17-25, Page No 408¶

In :
import math

#Initilization of variables
Mm=70 #kg
Mc=45 #kg
R=0.6 #m
g=9.8 #m/s**2
l=5 #m
# as theta=50 degrees,
sintheta=0.77

#Calculations
#T2 calculations except for v term in it as it cannot be declared as a number
T2=68.7 #without the v term in it
v=((g*Mm*l-g*Mc*l*sintheta)/T2)**0.5 #m/s

#Result
print'The speed is',round(v,2),"m/s"

The speed is 5.02 m/s


## Example 17.17-26, Page No 408¶

In :
import math

#The textbook has a typo in printing the question number
#Initilization of variables
W1=96.6 #lb
W2=128.8 #lb
v=8 #ft/s
g=32.2 #ft/s**2
# as theta=30 degrees,
sintheta=2**-1

#Calculations
#Initial KE of the system is T1=0
T2=(0.5*(W1/g)*v**2)+(0.5*(W2/g)*(v/2)**2) #ft-lb
#Work Done without s term
U=-(W1*sintheta)+W2*0.5
#S calculations
s=T2*U**-1 #ft

#Result
print'The block attains a speed of 8 ft/s in',round(s,2),"ft (up the plane)"

The block attains a speed of 8 ft/s in 7.95 ft (up the plane)


## Example 17.17-28, Page No 409¶

In :
# Initilization of variables
l=6 #m length of the cable
m=50 #kg mass of the cable
g=9.8 #m/s**2

import math
def integrand(x, a, b):
return 81.7*(6-x)
a=1
b=1

# Results
print'The word done is',round(Work),"N.m"
# The answer in textbook is off by 1 N.m

The word done is 1471.0 N.m


## Example 17.17-31, Page No 410¶

In :
# Initilization of variables
k=0.044 # spring constant
#x=0.300 #m length of compression from 450 to 150

import math
def integrand(x, a, b):
return 0.5*0.044*x**2
a=1
b=1

# Results
print'The word done on the balls is',round(W*10**-5,2),"N.m"

The word done on the balls is 1.98 N.m


## Example 17.17-32, Page No 410¶

In :
import math

#Initilization of variables
m=10 #kg
d=1.2 #m
g=9.8 #m/s**2

#Calculations
#Initilial KE is zero
#Final KE is(without v^2 term in it)
KE2=(3*4**-1)*10
#Work Done
U=m*g*d #N.m
#Velocity calculations
v=sqrt(U*KE2**-1) #m/s

#Result
print'The velocity is',round(v,2),"m/s"

The velocity is 3.96 m/s


## Example 17.17-33, Page No 411¶

In :
import math

#Initilization of variables
W=161 #lb
wa=150 #lb
wb=100 #lb
la=2 #ft
lb=4 #ft

#Calculations
#Work Done
T1=wb*lb-wa*la #ft-lb
#Final KE=zero
T2=0 #ft-lb
#Work Done on the system=T2-T1
#Hence the equation becomes
#50x-50x^2+100=0
#where
a=-50
b=50
c=100
#Solution
d=sqrt(b**2-4*a*c)
x1=(-b+d)/(2*a) #ft
x2=(-b-d)/(2*a) #ft

#Result
print'The stretch of the spring is',round(x2),"ft"
#Here even x1 could have been the solution,but the stretch in the string is elongation not compression hence x2 is the valid answer

The stretch of the spring is 2.0 ft


## Example 17.17-34, Page No 411¶

In :
import math

#Initilization of variables
I=100 #slug-ft**2
Mc=64.4 #lb
g=32.2 #ft/s**2

#Calculations
vb=2*w #ft/s
vc=0.5*w #ft/s
Mb=(0.5*I*w**2+0.5*(Mc/g)*vc**2+0.5*Mc*theta)/(2*theta-(0.5*vb**2*(1/g))) #lb

#Result
print'The weight of the block B is',round(Mb,1),"lb"

The weight of the block B is 90.6 lb


## Example 17.17-35, Page No 411¶

In :
import math

#Initilization of variables
Wa=96.6 #lb
Wb=128.8 #lb
g=32.2 #ft/s**2
I=12 #slug-ft**2
v=16 #ft/s
ratio=3**-1 #ratio of Sb/Sa
r=3#ft
va=6 #ft/s
vb=2 #ft/s

#Calculations
#Work Done without S in it
W=Wa-(ratio*Wb)
#System has zero KE initially and final KE is given by
T2=(0.5*(Wa/g)*va**2+0.5*I*w**2+0.5*(Wb/g)*vb**2) #ft-lb
#Distance Calculations
S=T2*W**-1 #ft

#Result
print'The distance through which A falls is',round(S,2),"ft"

The distance through which A falls is 1.6 ft


## Example 17.17-36, Page No 412¶

In :
import math

#initilization of variables
u=0.25 #coefficient of friction
k=2800 #N/m
x=0.075 #m
g=9.8 #m/s**2
m=7 #kg
# as theta=30 degrees,
sintheta=2**-1
costheta=(3**0.5)*2**-1

#Calculations
#Normal Reaction
N=g*m*costheta #N
#Frictional Force
Fr=u*N #N
#Component of force along the plane
F=g*m*sintheta #N
#Spring work is
W=0.5*k*x*x #N.m
s=(W+Fr*x-F*x)/(F-Fr) #m
S=(s*1000) #mm

#Result
print'The value of S is',round(S),"mm"

The value of S is 330.0 mm


## Example 17.17-37, Page No 412¶

In :
import math

#Initilization of variables
m=5 #kg
l=2 #m
k=10000 #N/m
x=0.1 #m
g=9.8 #m/s**2

#Calculations
drop=l+x #m mass drop length
#Work Done by Gravity
Wg=g*m*drop #N.m
#Work Done by Spring
Ws=0.5*k*x**2 #N.m
#Increase in KE is without v**2
KE=0.5*m #kg
#Velocity Calculations
v=sqrt((Wg-Ws)/KE) #m/s

#Result
print'The speed is',round(v,1),"m/s"

The speed is 4.6 m/s


## Example 17.17-38, Page No 412¶

In :
import math

#Initilization of variables
l=6 #ft
k=20 #lb/in
x=8 #in

#Calculations
#Work Done by Gravity
Wg=(l*12+x) #in without W
#Work Done by Spring
Ws=0.5*k*x**2 #in-lb
#Change in the kinetic energy is zero
W=Ws/Wg #lb

#Result
print'The weight is',round(W),"lb"

The weight is 8.0 lb


## Example 17.17-40, Page No 413¶

In :
import math

#Initilization of variables
W=8 #lb

#Calculations
#work done by the spring woithout k
Ws=0.5*((9*12**-1)**2-(12**-1)**2)
#Work done by gravity
Wg=W*(10.5*12**-1) #ft-lb
#Change in KE is zero
k=Wg/Ws #lb/ft

#Result
print'The value of k is',round(k,1),"lb/ft"

The value of k is 25.2 lb/ft


## Example 17.17-41, Page No 414¶

In :
import math

#Initilization of variables
Wc=100 #lb
r= 1 #ft
F=80 #lb
k=50 #lb/ft
s=6 #in
g=32.2 #ft/s**2

#Calculations
#Work done on the system
U=-0.5*k*(1)+F*(s*12**-1) #ft-lb
#Initial KE is zero
Vo=(U/(0.5*(Wc/g+0.5*(Wc/g)*r)))**0.5 #ft/s

#Result
print'The initial speed is',round(Vo,2),"ft/s"

The initial speed is 2.54 ft/s