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diff --git a/Principles_of_Physics_by_F.J.Bueche/Chapter5_2.ipynb b/Principles_of_Physics_by_F.J.Bueche/Chapter5_2.ipynb new file mode 100644 index 00000000..64ed1a47 --- /dev/null +++ b/Principles_of_Physics_by_F.J.Bueche/Chapter5_2.ipynb @@ -0,0 +1,618 @@ +{ + "cells": [ + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "# Chapter 05:Work and Energy" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.1:pg-159" + ] + }, + { + "cell_type": "code", + "execution_count": 1, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The work done is W= 0.0 Joules\n" + ] + } + ], + "source": [ + " import math #Example 5_1\n", + " \n", + " \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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.2:pg-159" + ] + }, + { + "cell_type": "code", + "execution_count": 2, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work done when lifting is W=mgh* 1.0 \n", + "\n", + "Work done when downing is W=mgh* -1.0 \n", + "\n" + ] + } + ], + "source": [ + " import math #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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.3:pg-160" + ] + }, + { + "cell_type": "code", + "execution_count": 3, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Work done is W= 100.0 Joules\n" + ] + } + ], + "source": [ + " import math #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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.4:pg-160" + ] + }, + { + "cell_type": "code", + "execution_count": 4, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Power developed P= 0.07882 hp\n" + ] + } + ], + "source": [ + " import math #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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.5:pg-161" + ] + }, + { + "cell_type": "code", + "execution_count": 5, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Average frictional force f= 4000.0 N\n" + ] + } + ], + "source": [ + " import math #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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.6:pg-162" + ] + }, + { + "cell_type": "code", + "execution_count": 8, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The speed of the car is vf= 14.1 meters/sec\n" + ] + } + ], + "source": [ + " import math #Example 5_6\n", + " \n", + "\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( vf,1),\" meters/sec\"\n" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.7:pg-163" + ] + }, + { + "cell_type": "code", + "execution_count": 9, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Tension in the rope is T= 396.0 N\n" + ] + } + ], + "source": [ + " import math #Example 5_7\n", + "\n", + "\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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.8:pg-163" + ] + }, + { + "cell_type": "code", + "execution_count": 10, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Frictional force required is f= 6000.0 N\n" + ] + } + ], + "source": [ + " import math #Example 5_8\n", + " \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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.9:pg-164" + ] + }, + { + "cell_type": "code", + "execution_count": 12, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The ball is moving with a speed of vf= 8.85 meters/sec\n" + ] + } + ], + "source": [ + " import math #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*math.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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.10:pg-164" + ] + }, + { + "cell_type": "code", + "execution_count": 13, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average frictional force f= 15.9 N\n" + ] + } + ], + "source": [ + " import math #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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.11:pg-165" + ] + }, + { + "cell_type": "code", + "execution_count": 14, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "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" + ] + } + ], + "source": [ + " import math #Example 5_11\n", + " \n", + " \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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.12:pg-166" + ] + }, + { + "cell_type": "code", + "execution_count": 15, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "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" + ] + } + ], + "source": [ + " import math #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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.13:pg-167" + ] + }, + { + "cell_type": "code", + "execution_count": 16, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "The average force required is f= 6553.0 N\n" + ] + } + ], + "source": [ + " import math #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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.14:pg-168" + ] + }, + { + "cell_type": "code", + "execution_count": 18, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "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" + ] + } + ], + "source": [ + " import math #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=math.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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.15:pg-169" + ] + }, + { + "cell_type": "code", + "execution_count": 19, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "Force required is F= 3313.0 N\n" + ] + } + ], + "source": [ + " import math #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" + ] + }, + { + "cell_type": "markdown", + "metadata": {}, + "source": [ + "## Ex5.16:pg-169" + ] + }, + { + "cell_type": "code", + "execution_count": 20, + "metadata": { + "collapsed": false + }, + "outputs": [ + { + "name": "stdout", + "output_type": "stream", + "text": [ + "IMA= 3.0 \n", + "\n", + "AMA= 2.5 \n", + "\n", + "Percentage of efficiency is 83.0 percent\n" + ] + } + ], + "source": [ + " import math #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" + ] + } + ], + "metadata": { + "kernelspec": { + "display_name": "Python 2", + "language": "python", + "name": "python2" + }, + "language_info": { + "codemirror_mode": { + "name": "ipython", + "version": 2 + }, + "file_extension": ".py", + "mimetype": "text/x-python", + "name": "python", + "nbconvert_exporter": "python", + "pygments_lexer": "ipython2", + "version": "2.7.11" + } + }, + "nbformat": 4, + "nbformat_minor": 0 +} |