# Chapter 05:Work and Energy¶

## Ex5.1:pg-159¶

In :
  import math  #Example 5_1

#To calculate the work done
Fs=8.0     #units in meters
W=Fs*round(math.cos(math.pi/2.0))     #units in Joules
print "The work done is W=",round(W)," Joules"

The work done is W= 0.0  Joules


## Ex5.2:pg-159¶

In :
  import math  #Example 5_2

#To calculate the work done when lifting object as well as lowering the object
Fs=1.0     #units in terms of Fs
theta=0     #units in degrees
W=Fs*math.cos(theta*math.pi/180)     #units in terms of m, g and h
print "Work done when lifting is W=mgh*",round(W),"\n"
theta=180.0     #units in degrees
W=Fs*math.cos(theta*math.pi/180)     #units in terms of m, g and h
print "Work done when downing is W=mgh*",round(W),"\n"

Work done when lifting is W=mgh* 1.0

Work done when downing is W=mgh* -1.0



## Ex5.3:pg-160¶

In :
  import math  #Example 5_3

#To find the work done by the pulling force
F=20.0     #units in Newtons
d=5     #units in meters
W=F*d     #units in joules
print "Work done is W=",round(W)," Joules"

Work done is W= 100.0  Joules


## Ex5.4:pg-160¶

In :
  import math  #Example 5_4

#To find out the power being developed in motor
m=200    #units on Kg
g=9.8     #units in meters/sec**2
Fy=m*g     #units in Newtons
vy=0.03     #units in meter/sec
P=Fy*vy     #units in Watts
P=P*(1.0/746)    #units in hp
print "Power developed P=",round(P,5)," hp"

Power developed P= 0.07882  hp


## Ex5.5:pg-161¶

In :
  import math  #Example 5_5

#To calculate the average frictional force developed
m=2000    #units in Kg
vf=20     #units in meters/sec
d=100    #units in meters
f=(0.5*m*vf**2)/d     #units in Newtons
print "Average frictional force f=",round(f)," N"

Average frictional force f= 4000.0  N


## Ex5.6:pg-162¶

In :
    import math  #Example 5_6

#To find out how fast the car is going
f=4000.0     #units in Newtons
s=50.0     #units in meters
theta=180.0     #units in degrees
m=2000.0      #units in Kg
v0=20     #units in meter/sec
vf=math.sqrt((2*((f*s*math.cos(theta*math.pi/180))+(0.5*m*v0**2)))/m)      #units in meter/sec
print "The speed of the car is vf=",round( vf,1)," meters/sec"

The speed of the car is vf= 14.1  meters/sec


## Ex5.7:pg-163¶

In :
  import math  #Example 5_7

#To find the required tension in the rope
m=40     #units in Kg
g=9.8     #units in meters/sec**2
theta=0    #units in degrees
vf=0.3     #units in meters/sec
s=0.5     #units in meters
T=round((m*g)+((0.5*m*vf**2)/(s*math.cos(theta*math.pi/180))))     #units in Newtons
print "Tension in the rope is T=",round(T)," N"

Tension in the rope is T= 396.0  N


## Ex5.8:pg-163¶

In :
  import math  #Example 5_8

#To calculate the frictional force
m=900.0    #units in Kg
v0=20.0     #units in meters/sec
s=30      #units in meters
f=(0.5*m*v0**2)/s      #units in Newtons
print "Frictional force required is f=",round(f)," N"

Frictional force required is f= 6000.0  N


## Ex5.9:pg-164¶

In :
  import math  #Example 5_9

#To find out how fast a a ball is going
m=3      #units in Kg
g=9.8     #units in meters/sec**2
hf=0     #units in meters
h0=4     #units in meters
vf=2*math.sqrt(((m*g*-(hf-h0))*0.5)/m)     #units in meters/sec
print "The ball is moving with a speed of vf=",round(vf,2)," meters/sec"

The ball is moving with a speed of vf= 8.85  meters/sec


## Ex5.10:pg-164¶

In :
  import math  #Example 5_10

#To calculate how large the average frictional force
a=9.8     #units in meters/sec**2
s=4     #units in meters
v=6      #units in meters/sec
m=3     #units on Kg
f=m*((a*s)-(0.5*v**2))/s     #units in Newtons
print "The average frictional force f=",round(f,1)," N"

The average frictional force f= 15.9  N


## Ex5.11:pg-165¶

In :
  import math  #Example 5_11

#To find out how fast a car is going at points B and C
m=300    #units in Kg
g=9.8    #units in meters/sec**2
hb_ha=10     #units in meters
f=20     #units in Newtons
s=60     #units in meters
vf=2*math.sqrt((0.5*((m*g*(hb_ha))-(f*s)))/m)     #units in meters/sec
print "The car is going at a speed of vf=",round(vf,1)," meters/sec at point B\n"
hc_ha=2     #units in meters
vf=2*math.sqrt((0.5*((m*g*(hc_ha))-(f*s)))/m)     #units in meters/sec
print "The car is going at a speed of vf=",round(vf,2)," meters/sec at point C\n"

The car is going at a speed of vf= 13.7  meters/sec at point B

The car is going at a speed of vf= 5.59  meters/sec at point C



## Ex5.12:pg-166¶

In :
  import math  #Example 5_12

#How far the average velocity and how far beyond B does the car goes
m=2000    #units in Kg
vb=5     #units in meters/sec
va=20      #units in meters/sec
hb_ha=8     #units in meters
g=9.8     #units in meters/sec**2
sab=100    #units in meters
f=-((0.5*m*(vb**2-va**2))+(m*g*(hb_ha)))/sab     #units in Newtons
print "Average frictional force is f=",round(f)," N\n"
Sbe=(0.5*m*vb**2)/f     #units in meters
print "The distance by which the car goes beyond is Sbe=",round(Sbe,1)," meters"
#In text book answer is printed wrong as f=2180 N but correct answer is f=2182N

Average frictional force is f= 2182.0  N

The distance by which the car goes beyond is Sbe= 11.5  meters


## Ex5.13:pg-167¶

In :
  import math  #Example 5_13

#To find out how large the force is required
m=2     #units in Kg
g=9.8     #units in meters/sec**2
hc_ha=10.03     #units in meters
sbc=0.030     #units in meters
f=(m*g*(hc_ha))/sbc     #units in Newtons
print "The average force required is f=",round(f)," N"
#In text book answer is printed wrong as f=6550 N correct answer is f=6552N

The average force required is f= 6553.0  N


## Ex5.14:pg-168¶

In :
  import math  #Example 5_14

#To find out how fast the pendulum is moving
#At point A
hb_ha=0.35    #units in Meters
g=9.8     #units in meters/sec**2
vb=math.sqrt((g*hb_ha)/0.5)     #units in meters/sec
print "The velocity of pendulum at point B is vb=",round(vb,2)," meters/sec\n"
print "From A to C hc=ha and Vc=Va=0 so Frictional force is Negligible at point C"

The velocity of pendulum at point B is vb= 2.62  meters/sec

From A to C hc=ha and Vc=Va=0 so Frictional force is Negligible at point C


## Ex5.15:pg-169¶

In :
  import math  #Example 5_15

#To find out how large a force is required
m=2000   #units in Kg
vf=15     #units in meters/sec
f1=500     #units in Newtons
F=((0.5*m*(vf**2))/80)+f1     #units in Newtons
print "Force required is F=",round(F)," N"
#In text book the answer is printed wrong as F=3300 N but the correct answer is 3312 N

Force required is F= 3313.0  N


## Ex5.16:pg-169¶

In :
  import math  #Example 5_16

#To find IMA AMA and Efficiency of the system
si=3.0
so=1
IMA=si/so
Fo=2000.0    #units in Newtons
Fi=800.0     #units in Newtons
AMA=Fo/Fi
effi=AMA/IMA*100
print "IMA=",round(IMA,2),"\n"
print "AMA=",round(AMA,2),"\n"
print "Percentage of efficiency is ",round(effi)," percent"

IMA= 3.0

AMA= 2.5

Percentage of efficiency is  83.0  percent