{
"metadata": {
"name": "",
"signature": "sha256:8f34af9d7ab4f4986ce62fe277146eddcd673716e93f679d03460223f2973812"
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"worksheets": [
{
"cells": [
{
"cell_type": "heading",
"level": 1,
"metadata": {},
"source": [
"Chapter 05:Work and Energy"
]
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.1:pg-159"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_1\n",
" \n",
"import math \n",
" #To calculate the work done\n",
"Fs=8.0 #units in meters\n",
"W=Fs*round(math.cos(math.pi/2.0)) #units in Joules\n",
"print \"The work done is W=\",round(W),\" Joules\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The work done is W= 0.0 Joules\n"
]
}
],
"prompt_number": 6
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.2:pg-159"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_2\n",
" \n",
" \n",
" #To calculate the work done when lifting object as well as lowering the object\n",
"Fs=1.0 #units in terms of Fs\n",
"theta=0 #units in degrees\n",
"W=Fs*math.cos(theta*math.pi/180) #units in terms of m, g and h\n",
"print \"Work done when lifting is W=mgh*\",round(W),\"\\n\"\n",
"theta=180.0 #units in degrees\n",
"W=Fs*math.cos(theta*math.pi/180) #units in terms of m, g and h\n",
"print \"Work done when downing is W=mgh*\",round(W),\"\\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Work done when lifting is W=mgh* 1.0 \n",
"\n",
"Work done when downing is W=mgh* -1.0 \n",
"\n"
]
}
],
"prompt_number": 8
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.3:pg-160"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_3\n",
" \n",
" \n",
" #To find the work done by the pulling force\n",
"F=20.0 #units in Newtons\n",
"d=5 #units in meters\n",
"W=F*d #units in joules\n",
"print \"Work done is W=\",round(W),\" Joules\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Work done is W= 100.0 Joules\n"
]
}
],
"prompt_number": 10
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.4:pg-160"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_4\n",
" \n",
" \n",
" #To find out the power being developed in motor\n",
"m=200 #units on Kg\n",
"g=9.8 #units in meters/sec**2\n",
"Fy=m*g #units in Newtons\n",
"vy=0.03 #units in meter/sec\n",
"P=Fy*vy #units in Watts\n",
"P=P*(1.0/746) #units in hp\n",
"print \"Power developed P=\",round(P,5),\" hp\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Power developed P= 0.07882 hp\n"
]
}
],
"prompt_number": 11
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.5:pg-161"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_5\n",
" \n",
" \n",
" #To calculate the average frictional force developed\n",
"m=2000 #units in Kg\n",
"vf=20 #units in meters/sec\n",
"d=100 #units in meters\n",
"f=(0.5*m*vf**2)/d #units in Newtons\n",
"print \"Average frictional force f=\",round(f),\" N\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Average frictional force f= 4000.0 N\n"
]
}
],
"prompt_number": 12
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.6:pg-162"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_6\n",
" \n",
"import math\n",
" #To find out how fast the car is going\n",
"f=4000.0 #units in Newtons\n",
"s=50.0 #units in meters\n",
"theta=180.0 #units in degrees\n",
"m=2000.0 #units in Kg\n",
"v0=20 #units in meter/sec\n",
"vf=math.sqrt((2*((f*s*math.cos(theta*math.pi/180))+(0.5*m*v0**2)))/m) #units in meter/sec\n",
"print \"The speed of the car is vf=\",round( ,1),\" meters/sec\",vf)\n"
],
"language": "python",
"metadata": {},
"outputs": []
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.7:pg-163"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_7\n",
"\n",
"import math\n",
" \n",
" #To find the required tension in the rope\n",
"m=40 #units in Kg\n",
"g=9.8 #units in meters/sec**2\n",
"theta=0 #units in degrees\n",
"vf=0.3 #units in meters/sec\n",
"s=0.5 #units in meters\n",
"T=round((m*g)+((0.5*m*vf**2)/(s*math.cos(theta*math.pi/180)))) #units in Newtons\n",
"print \"Tension in the rope is T=\",round(T),\" N\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Tension in the rope is T= 396.0 N\n"
]
}
],
"prompt_number": 14
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.8:pg-163"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_8\n",
"import math \n",
" \n",
" #To calculate the frictional force\n",
"m=900.0 #units in Kg\n",
"v0=20.0 #units in meters/sec\n",
"s=30 #units in meters\n",
"f=(0.5*m*v0**2)/s #units in Newtons\n",
"print \"Frictional force required is f=\",round(f),\" N\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Frictional force required is f= 6000.0 N\n"
]
}
],
"prompt_number": 15
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.9:pg-164"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_9\n",
" \n",
" \n",
" #To find out how fast a a ball is going\n",
"m=3 #units in Kg\n",
"g=9.8 #units in meters/sec**2\n",
"hf=0 #units in meters\n",
"h0=4 #units in meters\n",
"vf=2*sqrt(((m*g*-(hf-h0))*0.5)/m) #units in meters/sec\n",
"print \"The ball is moving with a speed of vf=\",round(vf,2),\" meters/sec\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The ball is moving with a speed of vf= 8.85 meters/sec\n"
]
}
],
"prompt_number": 16
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.10:pg-164"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_10\n",
" \n",
" \n",
" #To calculate how large the average frictional force\n",
"a=9.8 #units in meters/sec**2\n",
"s=4 #units in meters\n",
"v=6 #units in meters/sec\n",
"m=3 #units on Kg\n",
"f=m*((a*s)-(0.5*v**2))/s #units in Newtons\n",
"print \"The average frictional force f=\",round(f,1),\" N\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The average frictional force f= 15.9 N\n"
]
}
],
"prompt_number": 17
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.11:pg-165"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_11\n",
" \n",
"import math \n",
" #To find out how fast a car is going at points B and C\n",
"m=300 #units in Kg\n",
"g=9.8 #units in meters/sec**2\n",
"hb_ha=10 #units in meters\n",
"f=20 #units in Newtons\n",
"s=60 #units in meters\n",
"vf=2*math.sqrt((0.5*((m*g*(hb_ha))-(f*s)))/m) #units in meters/sec\n",
"print \"The car is going at a speed of vf=\",round(vf,1),\" meters/sec at point B\\n\"\n",
"hc_ha=2 #units in meters\n",
"vf=2*math.sqrt((0.5*((m*g*(hc_ha))-(f*s)))/m) #units in meters/sec\n",
"print \"The car is going at a speed of vf=\",round(vf,2),\" meters/sec at point C\\n\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The car is going at a speed of vf= 13.7 meters/sec at point B\n",
"\n",
"The car is going at a speed of vf= 5.59 meters/sec at point C\n",
"\n"
]
}
],
"prompt_number": 19
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.12:pg-166"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_12\n",
" \n",
" \n",
" #How far the average velocity and how far beyond B does the car goes\n",
"m=2000 #units in Kg\n",
"vb=5 #units in meters/sec\n",
"va=20 #units in meters/sec\n",
"hb_ha=8 #units in meters\n",
"g=9.8 #units in meters/sec**2\n",
"sab=100 #units in meters\n",
"f=-((0.5*m*(vb**2-va**2))+(m*g*(hb_ha)))/sab #units in Newtons\n",
"print \"Average frictional force is f=\",round(f),\" N\\n\"\n",
"Sbe=(0.5*m*vb**2)/f #units in meters\n",
"print \"The distance by which the car goes beyond is Sbe=\",round(Sbe,1),\" meters\"\n",
" #In text book answer is printed wrong as f=2180 N but correct answer is f=2182N\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Average frictional force is f= 2182.0 N\n",
"\n",
"The distance by which the car goes beyond is Sbe= 11.5 meters\n"
]
}
],
"prompt_number": 20
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.13:pg-167"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_13\n",
" \n",
" \n",
" #To find out how large the force is required\n",
"m=2 #units in Kg\n",
"g=9.8 #units in meters/sec**2\n",
"hc_ha=10.03 #units in meters\n",
"sbc=0.030 #units in meters\n",
"f=(m*g*(hc_ha))/sbc #units in Newtons\n",
"print \"The average force required is f=\",round(f),\" N\"\n",
" #In text book answer is printed wrong as f=6550 N correct answer is f=6552N\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The average force required is f= 6553.0 N\n"
]
}
],
"prompt_number": 21
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.14:pg-168"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_14\n",
" \n",
" \n",
" #To find out how fast the pendulum is moving \n",
" #At point A\n",
"hb_ha=0.35 #units in Meters\n",
"g=9.8 #units in meters/sec**2\n",
"vb=sqrt((g*hb_ha)/0.5) #units in meters/sec\n",
"print \"The velocity of pendulum at point B is vb=\",round(vb,2),\" meters/sec\\n\"\n",
"print \"From A to C hc=ha and Vc=Va=0 so Frictional force is Negligible at point C\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"The velocity of pendulum at point B is vb= 2.62 meters/sec\n",
"\n",
"From A to C hc=ha and Vc=Va=0 so Frictional force is Negligible at point C\n"
]
}
],
"prompt_number": 23
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.15:pg-169"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_15\n",
" \n",
" \n",
" #To find out how large a force is required\n",
"m=2000 #units in Kg\n",
"vf=15 #units in meters/sec\n",
"f1=500 #units in Newtons\n",
"F=((0.5*m*(vf**2))/80)+f1 #units in Newtons\n",
"print \"Force required is F=\",round(F),\" N\"\n",
" #In text book the answer is printed wrong as F=3300 N but the correct answer is 3312 N\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"Force required is F= 3313.0 N\n"
]
}
],
"prompt_number": 24
},
{
"cell_type": "heading",
"level": 2,
"metadata": {},
"source": [
"Ex5.16:pg-169"
]
},
{
"cell_type": "code",
"collapsed": false,
"input": [
" #Example 5_16\n",
" \n",
" \n",
" #To find IMA AMA and Efficiency of the system\n",
"si=3.0\n",
"so=1\n",
"IMA=si/so\n",
"Fo=2000.0 #units in Newtons\n",
"Fi=800.0 #units in Newtons\n",
"AMA=Fo/Fi\n",
"effi=AMA/IMA*100\n",
"print \"IMA=\",round(IMA,2),\"\\n\"\n",
"print \"AMA=\",round(AMA,2),\"\\n\"\n",
"print \"Percentage of efficiency is \",round(effi),\" percent\"\n"
],
"language": "python",
"metadata": {},
"outputs": [
{
"output_type": "stream",
"stream": "stdout",
"text": [
"IMA= 3.0 \n",
"\n",
"AMA= 2.5 \n",
"\n",
"Percentage of efficiency is 83.0 percent\n"
]
}
],
"prompt_number": 25
}
],
"metadata": {}
}
]
}