• # question_answer Two blocks of masses M = 5 kg and m = 3 kg are  placed on a horizontal surface as shown in the figure. The coefficient of friction between the blocks is 0.5 and that between the block M and the horizontal surface is 0.7. What is the maximum horizontal force F that can be applied to block M so that the two blocks move without slipping? Take $g=10\,m{{s}^{-2}}.$ A)  4 N                            B) $16\,N$C)  24 N              D)  96 N

The force of friction between the block m and the block$M={{\mu }_{1}}\,mg.$Force of friction between block M and the horizontal surface$={{\mu }_{2}}(M+m)g.$As there is no slipping between the blocks, so the entire system can be considered as a single system. Invoking Newton's second law on composite system. Thus,$F=(M+m)a+{{\mu }_{2}}(M+m)g$           (1) Now, since the force on block m is ${{\mu }_{1}}mg,$ its acceleration is $a=\frac{{{\mu }_{1}}mg}{m}={{\mu }_{1}}g$ Using Eqs. (1) and (2), $F={{\mu }_{1}}(M+m)g+{{\mu }_{2}}(M+m)g$$=({{\mu }_{1}}+{{\mu }_{2}})(M+m)g$ $=(0.5+0.7)\times (5+3)\times 10=96\,N.$ Hence, the correction option is [d].