[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]

[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]

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