Question #a5256

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Answer 1 · 2016-11-27T01:59:26

given

Let

$a \to downward acceleration of M$ $a-> "downward acceleration of M "$

$T \to tension on rope$ $T-> "tension on rope "$

$g \to acceleration due to gravity = 9.8 m s^{- 2}$ $g-> " acceleration due to gravity "=9.8ms^-2$

Considering the forces on 9 kg block we can write

$T-mu12g=9a...........(1)$

Considering the forces on $M=5kg$ block we can write

$5g-T=5a...........(2)$

Adding (1) and (2) we get

$(5g-mu12g)=14a$

$a=(5g-mu12g)/14=((5-0.3xx12)9.8)/14=0.98ms^-2$

Answer 2 · 2017-04-03T13:40:54

$a=1.61" "m/s^2$

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$m_1=12 " "kg$

$m_2=9" "kg$

$M=5 " "kg$

$mu_k=0.30$

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$"Total act on OB surface is "F=m_1g+m_2g$

$"Reaction Force of OB surface is " F_a=-F=-(m_1g+m_2g)$

$"Effect of m1 mass on G H surface is "m_1g-F=m_1g-(m_1g+m_2g)$

$N=-m_2g$

$f_s=mu_k*N" , "f_s=0.3*9*9.81=26.487" "N$

$a=(Mg-f_s)/(M+m_2$

$a=(5*9.81-26.487)/(5+9)$

$a=(49.05-26.487)/14$

$a=1.61" "m/s^2$