Question

Question #a5256

Answer 1

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`

Answer 2

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

Explanation:

`m_1=12 " "kg`

`m_2=9" "kg`

`M=5 " "kg`

`mu_k=0.30`

`"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`