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Title: Another Pulley ProblemPost by: Bitupon on December 14, 2005, 03:56:15 pm
[size=18:825bcc8bd7][b:825bcc8bd7]Two blocks A and B of mass 1kg and 2kg respectively are connected by a string, passing over a light frictionless pulley. Both the blocks are resting on a horizontal floor and the pulley is held such that string remains just taut. At moment t=0 ,a force F=20t N starts acting on the pulley along vertically upward direction as shown in the figure. Calculate
(a) Velocity of A and B when loses contact with the floor. (b) Height raised by the pulley upto that instant and (c) Work done by the force upto that instant (Consider g=10m per sec sq.)[/b:825bcc8bd7] [/size:825bcc8bd7] (http://img208.imageshack.us/img208/2625/slaws0sj.jpg) [size=24:825bcc8bd7][color=red:825bcc8bd7]Solution: (a) Let T be the tension in the string . 2T=20t Or, T=10t N Let the block A loses its contact with the floor at time t=t¡¦. This happens when the tension in the string becomes equal to the weight of A.Thus, T=mg Or,10t¡¦=10 Or, t¡¦=1sec Similarly for block B,we have 10 t¡¦¡¦=20 Or, t¡¦¡¦=2sec i.e. the block B loses contact after 2secs. For block A , at time t such that t¡¦<t, let a be the acceleration in upward direction. Then 10t-10=a=(dv/dt) Or, dv=10(t-1)dt Integrating we get, v=5(t^2)-10t+5 At t=2sec , v=5m/s Am I correct so far? How to find the rest of the questions? Please help.[/color:825bcc8bd7] [/size:825bcc8bd7] Title: topic315Post by: Fu-Kwun Hwang on December 28, 2005, 12:46:13 pm
When the force F is large enough both blocks will be lifted, but if the force is not large enough , the heavy one will be supported by the base.
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