Help solve question 24 Share with your friends Share 0 Sanjay Upadhyay answered this Dear Student, 1) In first case when spring 2 is cut just after cutting the spring force mg act on A in downwaed direction so a=mgm=g (down ward) and force mg will act on B in upward direction so b=mg2m=g2 (upward) option (B) Now in second case when spring 1 is cut just after cutting the spring force 3mg act on B in downwaed direction so b=3mg2m=32g (downward) and force on A is zero as due to inertia of spring it will experience no force so a=02m=0option( A )2)In first case accelaretion (a)=net forcetotal mass=2mg-2mg4m=0let tension in the sring is T and spring constant is Kthen T-2mg=2ma=0T=2mgso Kx1=T=2mg x1=2mgKin second caseaccelaretion (a)=net forcetotal mass=3mg-2mg5m=g5let tension in the sring is T and spring constant is Kthen free body diagram of 2mT-2mg=2mg5T=12mg/5so Kx2=T=12mg/5 x2=12mg5Kin third caseaccelaretion (a)=net forcetotal mass=2mg-mg3m=g3let tension in the sring is T and spring constant is Kthen free body diagram of 2m2mg-T=2mg/3T=4mg/3so Kx3=T=4mg/3 x3=4mg3Kform here you can seex2>x1>x3 Regard 0 View Full Answer
