Consider the situatioin shown in figure. The system is released from rest and the block of mass 1.0 kg is found to have a speed 0.3 m/s ater it has descended thrugh a distance of 1 m. find the coefficient of kinetic friction between the block and the table.

Answer

A, B

Explanation

`m_1=4kg, m^2 =1kg, `
`V_2 =0.3m/sec`
`V_1 =2xx(0.3)=0.6m/sec`
(v_1=2x_2`m in this system) `h=1 m=height descent travelled by 1 kg block
S=2xx1=2m`
=distance travelled by 4kg block `u=0`
Applying change in K.E. =work done (for the system) `
(1/2)m_1v_1^2+(1/2)m_2v_2^2=(-muR)S+m_2gh`
[R=4g=40N]`

`rarr 1/2xx4xx(0.36)+1/2xx1xx(0.09)`
` =-mu40xx2+1xx40xx1`
`rarr 0.72+0.045=-80mu+10`
`rarr mu=9.236/80=0.12`