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//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
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