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Let m1=1 kg, m2=2kg and m3=3kg is figure...

Let `m_1=1 kg, m_2=2kg and m_3=3kg` is figure. Find the acceleration of `m_1, m_2 and m_3`. The string from the upper pulley to `m_1` is 20 cm when the system is released from rest. How long will it take before `m_1` strikes the pulley?

Text Solution

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From the free body diagram (. 2),
`T+m_1a-(m_1g+F)=0`
` rarr T=m_1g+F-m_1a`
From the feree body diagram ( 3)
`rarr T=m_2g+F+m_2a`
`rarr T= 5g+1-5a` ……..i. ltbr. ` rarr T=2g+1+2a`
From equations i. and ii. we have
`5g+1-5a=2g+1+2a`
` rarr 3g-7a=0
` rarr 7a=3g`
`rarra= (3g)/7=29.4/7`
`4.2 m/s^2 [g= 9.8 m/s^2]`
Hence acceleration of block is `4.2 m/s^2`.
After the string breaks, `m_1` moves downward with force F acting downward, then,
` m_1 a=f+m_1g` ltbr. `where, foerce = 1 N and`
`acceleration = 1/5 =0.2 m/s^2 `
`:. m_1 a=(1+5g)`
So, `acceleration = Force/ Mass`
` =(5(g+0.2))/5`
`=(g+0.2) m/s^2`.
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