{ "metadata": { "name": "", "signature": "sha256:8f34af9d7ab4f4986ce62fe277146eddcd673716e93f679d03460223f2973812" }, "nbformat": 3, "nbformat_minor": 0, "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": {} } ] }