# Chapter 3 Potential energy¶

## Example 3.1 page no 35¶

In [3]:
#given
m=0.04                                        #Mass of stone in kg
vi=25                                         #Initial velocity in m/s
vf=0                                          #Final velocity in m/s
yi=0                                          #Initial height in m

#Calculations
import math
Ui=(m*9.81*yi)
Ki=(1/2.0)*m*vi**2
Etotal=(Ui+Ki)
h=(Etotal/(m*9.8))
#when the stone is at (2/3)h, total energy is again same
v=math.sqrt((Etotal-(m*9.8*(2/3.0)*h))/((1/2.0)*m))

#Output
print"Maximum height it will reach is ",round(h,1),"m"
print"velocity when it is at the two-third of its maximum height is ",round(v,2),"m/s"

Maximum height it will reach is  31.9 m
velocity when it is at the two-third of its maximum height is  14.43 m/s


## Example 3.2 page no 36¶

In [4]:
#given
m=0.5                                       #Mass of the sphere in kg
vi=100                                      #Initial velocity in m/s
vf=20                                       #Final velocity in m/s

#Calculations
h=(vi**2-vf**2)/(2.0*9.8)
PE=(m*9.8*h)

#Calculations
print"Potential energy of the sphere is ",PE,"J"

Potential energy of the sphere is  2400.0 J


## Example 3.3 page no 36¶

In [5]:
#given
m=0.5                                    #Mass of the block in kg
x=0.05                                   #Distance to which block is pulled in m
k=300                                    #Force constant of the spring in N/m

#Calculations
import math
U=(1/2.0)*k*x**2
v=x*math.sqrt(k/m)

#Output
print"Potential energy of the block when spring is in stretched position is ",U,"J"
print"Velocity of the block when it passes through the equilibrium position is ",round(v,2)," m/s"

Potential energy of the block when spring is in stretched position is  0.375 J
Velocity of the block when it passes through the equilibrium position is  1.22  m/s


## Example 3.4 page no 37¶

In [7]:
#given
l=0.8                                            #Length of a simple pendulum in m
q=30                                             #Angle with the vertical through which the bob is released in degrees
q1=10                                            #Required angle in degrees

#Calculations
import math
v=math.sqrt(2*9.8*l*(math.cos(q1*3.14/180.0)-math.cos(q*3.14/180.0)))

#Output
print"Speed when the bob is at the angle of ",q1,"degrees with the vertical is ",round(v,2),"m/s"

Speed when the bob is at the angle of  10 degrees with the vertical is  1.36 m/s


## Example 3.5 page no 37¶

In [8]:
#given
m=(9.1*10**-31)                                     #Mass of the electron in kg
v=(3*10**8)                                         #Velocity of light in m/s
c=(1.6*10**-19)                                     #Charge of the electron in coloumbs

#Calculations
import math
Re=(m*v**2)/(c*10**6)
E=(Re/math.sqrt(1-0.9**2))

#Output
print"Rest energy of the electron is ",round(Re,3),"MeV"
print"Total energy is ",round(E,2),"MeV"

Rest energy of the electron is  0.512 MeV
Total energy is  1.17 MeV