Solve the previous problem if the firction coefficient between the 2.0 kg block and the plane below it is 0.5 and the plane below the 4.0 kg block is frictionless.

Answer

A, B

Explanation

`m_1=4, m_2=2kg`
FricitoN/Al coefficient between 2 kg block
and surface =0.5
R=10 cm=0.1m`
`I=0.5kg-m^2`
`-m_1gsintheta-T_1`
=m_1a`…..1
`-T_2-(m_2gsintheta+mu m_2gcostheta)`
`=m_2a`…2
`(T_1-T_2)=I_a/r^2`……3
Adding equation 1 and 2 we will get
`m_1gsintheta-(m_2gsintheta-mu mg costheta)+(T_2-T_1)`
`=m_a+m_2a`
`=4xx9.8xx91/sqrt2)
`{2xx9.8xx(1/sqrt2)+0.5xx2xx9.8xx(1/sqrt2)}`
`=(4+2+05/0.01)a`
`rarr 37.80-(13.90+6.95)=56a`
`rarr 27.8-20.3=56a`
`rarr a=7/56=0.125m/s^2`