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In the arrangement shown in fig. m(1)=1k...

In the arrangement shown in fig. `m_(1)=1kg, m_(2)=2 kg`. Pulleys are massless and strings are light. For what value of M, the mass `m_(1)` moves with constant velocity.

Text Solution

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Mass `m_(1)` moves with constant velocity if tension in the lower string.
`T_(1)=m_(1)g=(1)(10)=10N` …(i)
Tension in the upper string
`T_(2)=2T_(1)=20N` …(ii)
Acceleration of block M is, therefore,
`a=(T_2)/(M)=(20)/(M)` …(iii)
This is also the acceleration of pulley 2.

The absolute acceleration of amss `m_(1)` is zero. Thus, the acceleration of `m_(1)` relative to pulley 2 is a upward or acceleration of `m_(2)` with respect to pulley 2 is a downward, Drawing free-body diagram of `m_(2)` with respect to pulley 2.
Equation of motion gives
`20-(40)/(M)-10 = 2a=(40)/(M)`
solving this we get, `M=8 kg`.
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