The two blocks are connected by a light inextensible cord,which passes around sm
ID: 1816246 • Letter: T
Question
The two blocks are connected by a light inextensible cord,which passes around small massless pulleys as shown below. If block B is pulled down 500mm from the equilibrium position and released from rest, determine its speed when itreturns to the equilibrium position.The two blocks are connected by a light inextensible cord,which passes around small massless pulleys as shown below. If block B is pulled down 500mm from the equilibrium position and released from rest, determine its speed when itreturns to the equilibrium position.
Explanation / Answer
Fya = 2T - mag - kx = 0 Fyb = T - mbg = 0 T = (10)(9.81) = 98.1 N x = (2T - mag)/k x = (2*98.1 - 2*9.81)/k = 0.22 m At position one, h for 'b' is 0.5, h for 'a' is 0.25 + initial 0.22that is is stretched at equilibrium. T1 + U1 = T2 + U2 T1 = 0 U1 = magh - mbgh +0.5kx2 = (2)(9.81)(0.22+0.25) - (10)(9.81)(0.5) +0.5(800)(0.22+0.25)2 = 48.5314 U2 = magh + 0.5kx2 =(2)(9.81)(0.22) + 0.5(800)(0.22)2 = 23.6764 T2 = 0.5ma(Vb/2)2 +0.5mbVb2 =5.25Vb2 T2 = U1 - U2 5.25Vb2 = 48.5314 - 23.6764 Vb = 4.73 = 2.17 m/s
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