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Two blocks of masses 5kg and 2 kg are in...

Two blocks of masses 5kg and 2 kg are initially at rest on the floor. They are connected by a light string. Passing over a light frictionless pulley. An upwards force F is applied on the pulley and maintained at a constant value. Calculate the acceleration `a_(1)` and `a_(2)` of the 5-kg and 2-kg masses, respectively, when F is (take `g=10ms^(-2)` )
(a) 30N, (b) 60N , (c) 140N
.

Text Solution

Verified by Experts

Apart from the constraint that the string is unstratchable, the additional constraint is that neither of the mass can go downward. So the block will be lifted only when the tension of the string exceeds the gravitational pull on them.

(a) When `f=30N`: Considering the free body diagram of the pulley
`30-2T=0 ("pulley is massless")`
or `T=15N`
So, tension is less than gravitational pull on both the blocks So, no acceleration is produced in them. Therefore,
`a_(1)=a_(2)=0`
(b) When `F=60N`: Now `60-2T=0` or `T=30N`
So, the 5-kg wieght will not be lifted but the 2kg weight will. For acceleration of 2-kg block, writing equation of motion
`30-20=2 xx a_(2)` or `a_(2)=5ms^(-2)`
thus, `a_(1)=0` and `a_(2)=5ms^(-2)`
(c) When `F=140N`: Now `140-2T=0 `ot `T=70N`
So both the weight are lifted. Writing equation of motion
For 5-kg block:`70-50=5a_(1) `or `a_(1)=4ms^(-2)`
For 2-kg block: `70-20=2a_(2)` or `a_(2)=25 ms^(-2)`.
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Knowledge Check

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