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Question Number 136930 by I want to learn more last updated on 27/Mar/21

Commented by I want to learn more last updated on 27/Mar/21

Commented by Olaf last updated on 28/Mar/21

(i) W = (1/2)Kx^2   K = ((2W)/x^2 ) = ((2×10)/(0,05^2 )) = 8000 Nm^(−1)   (ii) F = Kx = 8000×0,05 = 400 N  (iii) F = ma ⇒ a = (F/m) = ((400)/2) = 200 ms^(−2)   (iv) V_P  = at = 200×0,25 = 50 ms^(−1)   M_P V_P  + M_Q V_Q  = (M_P +M_Q )V  2×50+4×0 = (2+4)V  V = 16,7 ms^(−1)

(i)W=12Kx2K=2Wx2=2×100,052=8000Nm1(ii)F=Kx=8000×0,05=400N(iii)F=maa=Fm=4002=200ms2(iv)VP=at=200×0,25=50ms1MPVP+MQVQ=(MP+MQ)V2×50+4×0=(2+4)VV=16,7ms1

Commented by mr W last updated on 28/Mar/21

all is correct except (iv), since the  block P doesn′t move with constant  acceleration after release.  period T=2π(√(m/k))=2π(√(2/(8000)))≈0.1 s  for the block P to return back to its  initial position it needs (T/4)≈0.025 s.  at this position the spring is   completely relaxed and losses   connection with the block P which then  moves with a constant velocity v_P :  v_P =(√((2W)/m))=(√((2×10)/2))=(√(10)) m/s  also at time t=0.25 s, the block P has  this velocity till it collides with block  Q. the velocity after collision is  v_(P+Q) =((M_P v_P )/(M_P +M_Q ))=((2×(√(10)))/(2+4))=((√(10))/3)≈1.05 m/s

alliscorrectexcept(iv),sincetheblockPdoesntmovewithconstantaccelerationafterrelease.periodT=2πmk=2π280000.1sfortheblockPtoreturnbacktoitsinitialpositionitneedsT40.025s.atthispositionthespringiscompletelyrelaxedandlossesconnectionwiththeblockPwhichthenmoveswithaconstantvelocityvP:vP=2Wm=2×102=10m/salsoattimet=0.25s,theblockPhasthisvelocitytillitcollideswithblockQ.thevelocityaftercollisionisvP+Q=MPvPMP+MQ=2×102+4=1031.05m/s

Commented by I want to learn more last updated on 28/Mar/21

I really appreciate sirs. God bless you.

Ireallyappreciatesirs.Godblessyou.

Commented by Olaf last updated on 28/Mar/21

I uninstall this app.  Good continuation sir.

Iuninstallthisapp.Goodcontinuationsir.

Commented by Tawa11 last updated on 14/Sep/21

nice

nice

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