Question #a5256

2 Answers
Nov 27, 2016

given

Let

#a-> "downward acceleration of M "#

#T-> "tension on rope "#

#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#

Apr 3, 2017

Answer:

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

Explanation:

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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#