//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