A massive platform of mass M is moving with speed `v=6ms^(-1)` At t = 0, at body of mass `m(m lt lt M)` is gently placed on the platform. If c-efficient of friction between the body and platform is `mu=0.3 and g=10 m//s^(2)`, then

Answer

6

Explanation

since, `M gt gt m`,the velocity of M remains unchanged after m is placed on to it. Acceleration of m is:
`a = (mu mg)/m = mug `
Initial relative acceleration
`a_(mM) = a_(mE) - a_(ME) = a- 0 = a`
and initial relative velocity,
`mu_(mM) = mu(mE) + mu_(ME) = 0 -v = -v`
Hence,
`S_(mM) = v^(2)/(2mug) = (6)^(2)/(2 xx 0.3 xx 10 = 6 m`