# Chapter 1 - Introduction and basic concepts¶

## Example: 1.1 Page: 4¶

In [22]:
from __future__ import division
# Soltion

print "Example: 1.1 - Page: 4\n\n"

#*****Data*****#
g_Earth = 9.83 #  [m/square s]
F_Earth = 800 #  [N]
g_Moon = 3.2 #  [m/square s]
#************#

# From the expression of force, the force on the man on the Eath's surface is given by:
# F = m*g_Earth
m = F_Earth/g_Earth #  [kg]

# On the moon, the weight of the mass is equal to the force acting on the mass on the moon and is given by
F_Moon = m*g_Moon #  [N]

print "Weight of the man on the moon is %0.2f N\n"%(F_Moon)#

Example: 1.1 - Page: 4

Weight of the man on the moon is 260.43 N



## Example: 1.2 Page: 5¶

In [1]:
# Solution

print "Example: 1.2 - Page: 5\n\n"

#*****Data*****#
m1 = 1.5 #  [mass of the body, kg]
m2 = 6*10**(24) #  [mass of the Earth, kg]
G = 6.672*10**(-11) #  [N.square m/square.kg]
r = 6000*10**(3) #  [m]
#************#

# According to Newton's universal law of gravity:
F = G*m1*m2/r**2 #  [N]
print "Gravitational force on the body is %.2f N\n"%(F)#

Example: 1.2 - Page: 5

Gravitational force on the body is 16.68 N



## Example: 1.3 Page: 5¶

In [2]:
# Solution

print "Example: 1.3 - Page: 5\n\n"

#*****Data*****#
r_Moon = 0.3 #  [km]
r_Earth = 1 #  [km]
m2 = 1 #  [mass of body, kg]
mMoon_By_mEarth = 0.013 #  [kg/kg]
#***************#

# According to the Newton's universal law of gravitation:
Fe_By_Fm = (1/mMoon_By_mEarth)*(r_Moon/r_Earth)**2#
print "Mass of 1 kg will weigh %.2f kg on moon\n"%(Fe_By_Fm)#

Example: 1.3 - Page: 5

Mass of 1 kg will weigh 6.92 kg on moon



## Example: 1.4 Page: 6¶

In [3]:
# Solution

print "Example: 1.4 - Page: 6\n\n"

#*****Data*****#
h = 40 #  [cm]
density = 14.02 #  [g/cubic cm]
g = 9.792 #  [m/square s]
#*************#

P = h*density*g/1000 #  [N/square cm]
P = P*10 #  [kPa]

print "The absolute pressure is %.3f kPa\n"%(P)#

Example: 1.4 - Page: 6

The absolute pressure is 54.914 kPa



## Example: 1.5 Page: 7¶

In [4]:
# Solution

print "Example: 1.5 - Page: 7\n\n"

#*****Data*****#
Patm = 112 #  [kPa]
density = 1200 #  [kg/cubic m]
g = 9.81 #  [m/sqaure s]
h = 0.62 #  [m]
#**************#

P = Patm + (density*g*h/1000) #  [kPa]

print "The absolute pressure within the container is %.3f kPa\n"%(P)#

Example: 1.5 - Page: 7

The absolute pressure within the container is 119.299 kPa



## Example: 1.6 Page: 9¶

In [5]:
# Solution

print "Example: 1.6 - Page: 9\n\n"

#*****Data*****#
F = 150 #  [N]
Displacement = 10 #  [m]
#**************#

W = F*Displacement #  [J]

print "Work done by the system is %d J\n"%(W)#

Example: 1.6 - Page: 9

Work done by the system is 1500 J



## Example: 1.7 Page: 9¶

In [6]:
# Solution

print "Example: 1.7 - Page: 9\n\n"

#*****Data*****#
P = 560*10**3 #  [Pa]
Vinit = 3 #  [cubic m]
Vfinal = 5 #  [cubic m]
Wext = 210*10**3 #  [J]
#*************#

W = P*(Vfinal - Vinit) #  [J]
# Again the system receives 210 kJ of work from the external agent.
W = W - Wext #  [J]

print "Actual Work done by the system is %.1e J\n"%(W)#

Example: 1.7 - Page: 9

Actual Work done by the system is 9.1e+05 J



## Example: 1.8 Page: 11¶

In [7]:
# Solution

print "Example: 1.8 - Page: 11\n\n"

#*****Data*****#
g = 9.81 #  [m/square s]
Z = 100 # [m]
#***************#

# Basis: 1 kg of water
m = 1 #  [kg]
Ep = m*g*Z #  [J]
print "Change in potential Energy is %d J\n"%(Ep)

Example: 1.8 - Page: 11

Change in potential Energy is 981 J



## Example: 1.9 Page: 11¶

In [8]:
from __future__ import division
# Solution

print "Example: 1.9 - Page: 11\n\n"

#*****Data*****#
m = 15# # [kg]
g = 9.81 #  [m/square s]
V1 = 0 #  [m/square s]
Z1 = 12 #  [m]
Z2 = 0 #  [m]
#***************#

# At initial condition, V1 = 0, so kinetic energy is zero.
# At final condition, Z2 = 0, so potential energy is zero.
# Ep1 + Ek1 = Ep2 + Ek2
#deff('[y] = f(V2)','y = ((1/2)*m*V1**2) + (m*g*Z1) - (((1/2)*m*V2**2) + (m*g*Z2))')#
def f(V2):
y = ((1/2)*m*V1**2) + (m*g*Z1) - (((1/2)*m*V2**2) + (m*g*Z2))
return y
from scipy.optimize import fsolve
V2 = fsolve(f,7)#

print "The velocity of the metal block is %.2f m/s\n"%(V2)#

Ek2 = (1/2)*m*V2**2 #  [J]
print "The final Kinetic Energy is %.1f J\n"%(Ek2)#

Example: 1.9 - Page: 11

The velocity of the metal block is 15.34 m/s

The final Kinetic Energy is 1765.8 J



## Example: 1.10 Page: 12¶

In [2]:
from __future__ import division
# Solution

print "Example: 1.10 - Page: 12\n\n"

#*****Data*****#
m = 1200 #  [kg]
v1 = 10 #  [km/h]
v2 = 100 #  [km/h]
time = 1 #  [min]
#***************#

v1 = 10*1000/3600 #  [m/s]
v2 = 100*1000/3600 #  [m/s]
W = (1/2)*m*(v2**2 - v1**2) #  [J]
time = time*60 #  [s]
P = W/time #  [W]
print "Power required is %.2f kW\n"%(P/1000)#

Example: 1.10 - Page: 12

Power required is 7.64 kW



## Example: 1.11 Page: 13¶

In [32]:
from __future__ import division
from math import pi
print "Example: 1.11 - Page: 13\n\n"

#*****Data*****#
dia = 0.3 #  [m]
m = 100 #  [kg]
P_atm = 1.013*10**5 #  [N/square m]
g = 9.792 #  [m/square s]
#**************#

Area = (pi/4)*dia**2 #  [square m]
#Solution (a)(i)
# Force exerted by the atmosphere:
F_atm = P_atm*Area #  [N]
# Force exerted by piston & metal block:
F_mass = m*g #  [N]
# Total force acting upon the gas:
F = F_atm + F_mass #  [N]
print "Total Force eacting upon the gas is %.1f N\n"%(F)#

# Solution (a)(ii)
Pressure = F/Area #  [N/square m]
print "Pressure exerted is %.3f kPa\n\n"%(Pressure/1000)#

# Solution (b)
# The gas expands on application of heat, the volume of the gas goes on increasing and the piston moves upward.
Z = 0.5 #  [m]
# Work done due to expansion of gas:
W = F*Z #  [J]
print "Work due to expansion by the gas is %.3f kJ\n\n"%(W/1000)#

# Solution (c)
# Change in potential energy of piston and weight after expansion process:
Ep = m*g*Z #  [J]
print "Change in Potential Energy is %.1f J\n"%(Ep)#

Example: 1.11 - Page: 13

Total Force eacting upon the gas is 8139.7 N

Pressure exerted is 115.153 kPa

Work due to expansion by the gas is 4.070 kJ

Change in Potential Energy is 489.6 J



## Example: 1.12 Page: 24¶

In [3]:
from __future__ import division
print "Example: 1.12 - Page: 24\n\n"

# Solution

# The relation is:
# (C/5) = ((F - 32)/9)
# For C = F
C = - (32*5/4)# # [degree Celsius]
print "The temperature which has the same value on both the centigrade and Fahrenheit scales is %d degree Celsius or %d degree Fahrenheit\n"%(C,C)#

Example: 1.12 - Page: 24

The temperature which has the same value on both the centigrade and Fahrenheit scales is -40 degree Celsius or -40 degree Fahrenheit



## Example: 1.13 Page: 24¶

In [4]:
from __future__ import division
print "Example: 1.13 - Page: 24\n\n"

# Solution

#*****Data*****#
delta_T_C = 30 #  [OC]
#*************#

# The relation between the Kelvin temperature scale and the Celsius temperature scale:
# T(K) = T(OC) + 273.15
# Here, the temperature rise is to be expressed in terms of K, but the difference in temperature will be the same in the Kelvin and Celsius scales of temperature:
delta_T_K = delta_T_C #  [K]
print "The rise in temperature in the Kelvin scale is %d K\n"%(delta_T_K)#
# The emperical relationship between the Rankine and Kelvin scales is given by:
delta_T_R = 1.8*delta_T_K #  [R]
print "The rise in temperature in the Rankine scale is %d R\n"%(delta_T_R)#
delta_T_F = delta_T_R #  [OF]
print "The rise in temperature in the Fahrenheit scale is %d OF\n"%(delta_T_F)#

Example: 1.13 - Page: 24

The rise in temperature in the Kelvin scale is 30 K

The rise in temperature in the Rankine scale is 54 R

The rise in temperature in the Fahrenheit scale is 54 OF