A massive platform of mass M is moving with speed v=6ms−1. At. t=0, a body of mass m(m<

Idea The kinetic friction will act only if there is a relative motion between the two surfaces. In this question the relative motion between them will stop when their velocities become same and in this situation the force of friction between them will become zero.\displaystyle{\Sin{{c}}}{e},{M}{\rangle}{m},{t}{h}{e}{v}{e}{l}{o}{c}{i}{t}{y}{o}{f}{M}{r}{e}{m}{a}\in{s}{u}{n}{c}{h}{a}{n}\ge{d}{a}{f}{t}{e}{r}{m}{i}{s}{p}{l}{a}{c}{e}{d}{o}{n}\to{i}{t}.{A}{\mathcal{{e}}}\le{r}{a}{t}{i}{o}{n}{o}{f}{m}{i}{s}{a}{a}=μ{m}{g}{m}={m}{g}Initial relative acceleration$amM=amE-amE=a-0=a$and initial relative velocity,$vmM=vmE+vmE=0-v=-v$Hence, SmM=v22μg=(6)22×0.3×10=6m$TESTEd\ge$$Intheabovecasethef\text{or}ceact\in gontheaboveblockisonly\mathfrak{i}ction,$⇒ So amax for the upper block = μmgm=μg$\Rightarrow So\text{if}the\max i\mu mae\le rationoflowerblockis\mu >henonlythey\perp hco\underline{d}move\to \ge therotherwisek\in etic\mathfrak{i}ctionwillact.$