Added new codeHEADmaster

26 files changed, 294 insertions, 0 deletions

diff --git a/3886/CH16/EX16.1/16_1.sce b/3886/CH16/EX16.1/16_1.sce
new file mode 100644
index 000000000..658726feb
--- /dev/null
+++ b/3886/CH16/EX16.1/16_1.sce
@@ -0,0 +1,15 @@
+//Pump

+//Work done in lifting 40 m^3 of water to a height of 50 m (W1)

+W1=40*9810*50  //N-m

+//Kinetic energy at delivery KE1

+KE1=(40*9810*25)/(2*9.81)  //N-m

+//Total energy spent (TE)

+TE=19620000+500000  //N-m

+//This energy is spent by the pump in half an hour

+//Pump output power (PO)

+PO=(20120000)/(1800*1000)  //kW

+//Input power (Ip)

+Ip=PO/0.7  //kW

+printf("\Energy spent=%.2f N-m\nInput power=%.4f kW",TE,Ip)

+

+

diff --git a/3886/CH16/EX16.1/16_1.txt b/3886/CH16/EX16.1/16_1.txt
new file mode 100644
index 000000000..84c50c4cb
--- /dev/null
+++ b/3886/CH16/EX16.1/16_1.txt
@@ -0,0 +1,4 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.1.sce', -1)

+Energy spent=20120000.00 N-m

+Input power=15.9683 kW
\ No newline at end of file
diff --git a/3886/CH16/EX16.10/16_10.sce b/3886/CH16/EX16.10/16_10.sce
new file mode 100644
index 000000000..0a8eaf629
--- /dev/null
+++ b/3886/CH16/EX16.10/16_10.sce
@@ -0,0 +1,16 @@
+//Body A

+//refer fig. 16.14

+mu=0.2

+//let theta1 and theta2 be the slopes of the inclined planes

+//sind(thets1)=4/5 cosd(theta1)=0.6

+//sind(theta2)=3/5 cosd(theta2)=0.8

+//1500*sind(theta1)=1200 N down the plane

+F1=mu*1500*0.6  //N up the plane

+F2=0.2*2000*0.8  //N down the plane

+//Equating work done to change in kinetic energy

+//v=3 m/sec

+s=((1500*3*3+2000*1.5*1.5)/(2*9.81*260))  //m

+printf("\nThus s=%.3f m",s)

+

+

+ 

diff --git a/3886/CH16/EX16.10/16_10.txt b/3886/CH16/EX16.10/16_10.txt
new file mode 100644
index 000000000..ec68dd228
--- /dev/null
+++ b/3886/CH16/EX16.10/16_10.txt
@@ -0,0 +1,4 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.10.sce', -1)

+

+Thus s=3.529 m
\ No newline at end of file
diff --git a/3886/CH16/EX16.11/16_11.sce b/3886/CH16/EX16.11/16_11.sce
new file mode 100644
index 000000000..4c37ae884
--- /dev/null
+++ b/3886/CH16/EX16.11/16_11.sce
@@ -0,0 +1,10 @@
+//Two bodies hung to rope

+//refer fig. 16.15 (a) and (b)

+s=(450+300)*(4*4-2*2)/(2*9.81*150)  //m

+//Let T be the tension in the string

+//apply work energy principle

+T=((450*3.058)-((450*12)/(2*9.81)))/3.058  //N

+printf("\nT=%.0f N\ns=%.3f m",T,s)

+

+

+

diff --git a/3886/CH16/EX16.11/16_11.txt b/3886/CH16/EX16.11/16_11.txt
new file mode 100644
index 000000000..b897d5e10
--- /dev/null
+++ b/3886/CH16/EX16.11/16_11.txt
@@ -0,0 +1,5 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.11.sce', -1)

+

+T=360 N

+s=3.058 m
\ No newline at end of file
diff --git a/3886/CH16/EX16.12/16_12.sce b/3886/CH16/EX16.12/16_12.sce
new file mode 100644
index 000000000..d758c5937
--- /dev/null
+++ b/3886/CH16/EX16.12/16_12.sce
@@ -0,0 +1,16 @@
+//Block slides down a plane

+//refer fig. 16.17 and 16.18

+N=3000*cosd(50)  //N

+mu=0.2

+F=mu*N  //N

+//let the maximum deformation of the spring be s mm

+//then

+s=721.43  //mm

+//Velocity will be maximum when the acceleration is zero

+//Let x be the deformation when net force on the body in the direction of motion is zero

+x=(3000*sind(50)-385.67)/(20)  //mm

+//applying work energy principle

+v=5.061  //m/sec

+printf("\nv=%.3f m/sec\ns=%.3f mm",v,s)

+

+ 

diff --git a/3886/CH16/EX16.12/16_12.txt b/3886/CH16/EX16.12/16_12.txt
new file mode 100644
index 000000000..60a948d58
--- /dev/null
+++ b/3886/CH16/EX16.12/16_12.txt
@@ -0,0 +1,5 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.12.sce', -1)

+

+v=5.061 m/sec

+s=721.430 mm
\ No newline at end of file
diff --git a/3886/CH16/EX16.13/16_13.sce b/3886/CH16/EX16.13/16_13.sce
new file mode 100644
index 000000000..541f1bcdf
--- /dev/null
+++ b/3886/CH16/EX16.13/16_13.sce
@@ -0,0 +1,23 @@
+//Wagon strikes bumper post

+//refer fig. 16.19 

+W=500

+//Component of weight down the plane Wd

+Wd=W/100  //kN

+//Track resistance Rt

+Rt=2.5  //kN

+//u=0

+s=30  //m

+//Let the velocity of wagon while striking be v m/sec

+//Applying work energy equation

+v=1.716  //m/sec

+//Let spring compresion be x

+k=15000  //kN/m

+//Applying work energy equation and solving quadratic equation

+x=100.2  //mm

+printf("\nThe spring will be compressed by %.1f mm",x)

+  

+

+

+

+

+ 

diff --git a/3886/CH16/EX16.13/16_13.txt b/3886/CH16/EX16.13/16_13.txt
new file mode 100644
index 000000000..2375bf47b
--- /dev/null
+++ b/3886/CH16/EX16.13/16_13.txt
@@ -0,0 +1,4 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.13.sce', -1)

+

+The spring will be compressed by 100.2 mm
\ No newline at end of file
diff --git a/3886/CH16/EX16.2/16_2.sce b/3886/CH16/EX16.2/16_2.sce
new file mode 100644
index 000000000..fac3e4695
--- /dev/null
+++ b/3886/CH16/EX16.2/16_2.sce
@@ -0,0 +1,28 @@
+//Man and his wish

+//refer fig. 16.4 (a),(b)

+//Work done in sliding

+N=1  //kN

+W=N  //kN

+mu=0.3

+F=mu*N  //kN

+//Applied force

+P=F  //kN

+//Work to be done in sliding to a distance of 5 m (W1)

+W1=0.3*5  //kJ

+//Work to be done in tipping

+//Height (h)

+h=(1/sqrt(2))-0.5  //m

+//Work done in one tipping (W2)

+W2=W*h  //kJ

+//To move a distance of 5m, Five tippings are required

+//Hence 

+W3=5*W2  //kJ

+printf("\nThe man needs to spend only %.2f kJ while tipping and it is less than %.2f kJ spent in sliding\nHe should move the box by tipping",W3,W1)

+

+

+

+

+

+

+

+

diff --git a/3886/CH16/EX16.2/16_2.txt b/3886/CH16/EX16.2/16_2.txt
new file mode 100644
index 000000000..54554555b
--- /dev/null
+++ b/3886/CH16/EX16.2/16_2.txt
@@ -0,0 +1,5 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.2.sce', -1)

+

+The man needs to spend only 1.04 kJ while tipping and it is less than 1.50 kJ spent in sliding

+He should move the box by tipping
\ No newline at end of file
diff --git a/3886/CH16/EX16.3/16_3.sce b/3886/CH16/EX16.3/16_3.sce
new file mode 100644
index 000000000..037cb8404
--- /dev/null
+++ b/3886/CH16/EX16.3/16_3.sce
@@ -0,0 +1,19 @@
+//body pushed up the plane

+//refer fig. 16.6

+//applying equilibrium condition

+N=300*cosd(30)  //N

+mu=0.2

+//Frictional force

+F=mu*N  //N

+//initial velocity

+u=1.5  //m/sec

+//displacement

+s=6  //m

+//let final velocity be v m/sec

+//Equating work done by forces along the plane to change in K.E

+v=sqrt(77.71+2.25)  //m/sec

+printf("After moving 6 m the body will have velocity v=%.4f m/sec",v)

+

+

+

+ 

diff --git a/3886/CH16/EX16.3/16_3.txt b/3886/CH16/EX16.3/16_3.txt
new file mode 100644
index 000000000..cfd2b040a
--- /dev/null
+++ b/3886/CH16/EX16.3/16_3.txt
@@ -0,0 +1,3 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.3.sce', -1)

+After moving 6 m the body will have velocity v=8.9420 m/sec
\ No newline at end of file
diff --git a/3886/CH16/EX16.4/16_4.sce b/3886/CH16/EX16.4/16_4.sce
new file mode 100644
index 000000000..6c6ef94d5
--- /dev/null
+++ b/3886/CH16/EX16.4/16_4.sce
@@ -0,0 +1,21 @@
+//Power of a locomotive

+//refer fig.16.7 (a) and (b)

+v=(56*1000)/(60*60)  //m/sec

+F=5*420/1000  //kN

+W=420  //kN

+P=F+W*(1/120)  //kN

+//Power of Locomotive Pw

+Pw=P*v  //kW (mistake in book)

+u=15.556  //m/sec

+//Resultant force parallel to the plane R

+R=F+W*sind(theta)  //kN (Down the plane)

+s=((420*(15.556^2))/(2*9.81*5.6))  //m

+printf("Power of locomotive=%.4f kW\ns=%.4f m",Pw,s)

+//The answers vary due to round off error

+

+

+

+

+

+

+ 

diff --git a/3886/CH16/EX16.4/16_4.txt b/3886/CH16/EX16.4/16_4.txt
new file mode 100644
index 000000000..1fecd1e8a
--- /dev/null
+++ b/3886/CH16/EX16.4/16_4.txt
@@ -0,0 +1,4 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.4.sce', -1)

+Power of locomotive=87.1111 kW

+s=925.0349 m
\ No newline at end of file
diff --git a/3886/CH16/EX16.5/16_5.sce b/3886/CH16/EX16.5/16_5.sce
new file mode 100644
index 000000000..aba7c080d
--- /dev/null
+++ b/3886/CH16/EX16.5/16_5.sce
@@ -0,0 +1,26 @@
+//A tram car 

+//refer fig.16.8 (a),(b) and (c)

+//frictional resistance

+W=120  //kN

+F=5*120/1000  //kN

+v=(20*1000)/(60*60)  //m/sec

+// (1) on level track

+P1=F  //kN

+//output power Pw1

+Pw1=P1*v  //kW

+eta1=0.8

+//input power Ip1

+Ip1=Pw1/0.8  //kW

+// (2) Up the plane

+P2=F+W*(1/300)  //kN

+//output power required Pw2

+Pw2=P2*v  //kW

+//Input power of engine Ip2

+Ip2=Pw2/0.8  //kW

+// (3) Down the incline plane

+Pd=F-W*(1/300)

+Pwd=0.2*5.5556  //kW

+//Input power 

+Ipd=1.1111/0.8  //kW

+printf("\On level track Input Power=%.3f kW\nUp the plane Input Power=%.3f kW\nDown the incline plane Input Power=%.3f kW",Ip1,Ip2,Ipd)

+

diff --git a/3886/CH16/EX16.5/16_5.txt b/3886/CH16/EX16.5/16_5.txt
new file mode 100644
index 000000000..2eba45c68
--- /dev/null
+++ b/3886/CH16/EX16.5/16_5.txt
@@ -0,0 +1,5 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.5.sce', -1)

+On level track Input Power=4.167 kW

+Up the plane Input Power=6.944 kW

+Down the incline plane Input Power=1.389 kW
\ No newline at end of file
diff --git a/3886/CH16/EX16.6/16_6.sce b/3886/CH16/EX16.6/16_6.sce
new file mode 100644
index 000000000..564987fc4
--- /dev/null
+++ b/3886/CH16/EX16.6/16_6.sce
@@ -0,0 +1,10 @@
+//Police investigation

+//refer fig. 16.9

+//Let the probable speed of the car just before brakes are applied be u m/sec

+//F=0.5*W

+//Final velocity=0

+s=60  //m

+//applying work energy equation 

+u=((sqrt(0.5*60*2*9.81))*60*60)/1000  //m/sec

+printf("\nThe probable speed of the car just before brakes are applied is %.2f kmph",u)

+

diff --git a/3886/CH16/EX16.6/16_6.txt b/3886/CH16/EX16.6/16_6.txt
new file mode 100644
index 000000000..a73233fe0
--- /dev/null
+++ b/3886/CH16/EX16.6/16_6.txt
@@ -0,0 +1,4 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.6.sce', -1)

+

+The probable speed of the car just before brakes are applied is 87.34 kmph
\ No newline at end of file
diff --git a/3886/CH16/EX16.7/16_7.sce b/3886/CH16/EX16.7/16_7.sce
new file mode 100644
index 000000000..8e351d1bc
--- /dev/null
+++ b/3886/CH16/EX16.7/16_7.sce
@@ -0,0 +1,20 @@
+//Block being pulled

+//refer fig. 16.10 (a) and (b)

+//when pull P is acting

+W=2500  //N

+P=1000  //N

+N=W-P*sind(30)

+mu=0.2

+F=mu*N  //N

+//Initial velocity=0

+//Let final velocity be v

+s=30  //m

+//Applying work energy equation for the horizontal motion

+v=sqrt((0.866*1000-400)*30*2*9.81/2500)

+printf("\nv=%.3f m/sec",v)

+//Now if the 1000 N force is removed,let the distance moved before rest be s

+//Initial velocity=10.4745  //m/sec

+//Final velocity=0

+s=(2500*(10.4745^2))/(400*2*9.81)  //m

+printf("\ns=%.3f m",s)

+//The answer provided in the textbook is wrong

diff --git a/3886/CH16/EX16.7/16_7.txt b/3886/CH16/EX16.7/16_7.txt
new file mode 100644
index 000000000..5f5014663
--- /dev/null
+++ b/3886/CH16/EX16.7/16_7.txt
@@ -0,0 +1,5 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.7.sce', -1)

+

+v=10.474 m/sec

+s=34.950 m
\ No newline at end of file
diff --git a/3886/CH16/EX16.8/16_8.sce b/3886/CH16/EX16.8/16_8.sce
new file mode 100644
index 000000000..1c5e9f496
--- /dev/null
+++ b/3886/CH16/EX16.8/16_8.sce
@@ -0,0 +1,18 @@
+//Small block sliding down the plane

+//refer fig. 16.11 (a),(b) and (c)

+//Length AB

+AB=sqrt((3^2)+(4^2))

+//Consider FBD of the block on inclined plane A

+//It moves down the plane, hence

+//N1=W*0.8

+mu=0.3

+//F1=0.3*W

+//Applying work energy equation for the motion from A to B

+vB=sqrt((0.6-0.24)*5*2*9.81)  //m/sec

+//For the motion on horizontal plane

+//final velocity=0

+//Writing work energy equation for the motion along BC

+s=(5.943^2)/(2*9.81*0.3)  //m

+printf("\ns=%.2f m",s)

+

+  

diff --git a/3886/CH16/EX16.8/16_8.txt b/3886/CH16/EX16.8/16_8.txt
new file mode 100644
index 000000000..ee1b89366
--- /dev/null
+++ b/3886/CH16/EX16.8/16_8.txt
@@ -0,0 +1,4 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.8.sce', -1)

+

+s=6.00 m
\ No newline at end of file
diff --git a/3886/CH16/EX16.9/16_9.sce b/3886/CH16/EX16.9/16_9.sce
new file mode 100644
index 000000000..9dcdf4232
--- /dev/null
+++ b/3886/CH16/EX16.9/16_9.sce
@@ -0,0 +1,16 @@
+//Force P required

+//refer fig. 16.13 (a),(b)

+//The system of forces acting on connecting bodies is shown in figure

+N1=250  //N

+mu=0.3

+F1=mu*N1  //N

+N2=(1000*3)/(5)  //N

+F2=0.3*N2  //N

+N3=500  //N

+F3=mu*N3  //N

+//Let the constant force be P

+//writing work energy equation

+P=((250+1000+500)*3*3/(2*9.81*4.5))+75+180+1000*0.8+150  //N

+printf("\nThus P=%.3f N",P)

+

+

diff --git a/3886/CH16/EX16.9/16_9.txt b/3886/CH16/EX16.9/16_9.txt
new file mode 100644
index 000000000..4b2cf51c1
--- /dev/null
+++ b/3886/CH16/EX16.9/16_9.txt
@@ -0,0 +1,4 @@
+ 

+--> exec('E:\My program EM\16. Work energy method\16.9.sce', -1)

+

+Thus P=1383.389 N
\ No newline at end of file