A block of mass $$15kg$$ is placed over a frictionless horizontal surface. Another block of mass $$10kg$$ is placed over it, that is connected with a light string passing over two pulleys fastened to the $$15kg$$ block. A force $$80N$$ is applied horizontally to the free end of the string. Friction coefficient between two blocks is $$0.6$$. The portion of the string between $$10kg$$ block and the upper pulley is horizontal. Pulley, string and connecting rods are massless. $$(Take$$ $$g=10$$ $$m/s2$$ $$)The$$ magnitude of acceleration of the $$10kg$$ block is

Question

A block of mass $$15kg$$ is placed over a frictionless horizontal surface. Another block of mass $$10kg$$ is placed over it, that is connected with a light string passing over two pulleys fastened to the $$15kg$$ block. A force $$80N$$ is applied horizontally to the free end of the string. Friction coefficient between two blocks is $$0.6$$. The portion of the string between $$10kg$$ block and the upper pulley is horizontal. Pulley, string and connecting rods are massless. $$(Take$$ $$g=10$$ $$m/s2$$ $$)The$$ magnitude of acceleration of the $$10kg$$ block is

Infinity Learn · Accepted Answer

First, let us check upto what value of F, both blocks move together. Till friction becomes limiting, they will be moving together. Using the FBDs $$10kg$$ block will not slip over the $$15kg$$ block till acceleration of $$15kg$$ block becomes maximum as it is created only by friction force exerted by $$10kg$$  block on it.Let say both blocks have same acceleration.F−$$f10=f15$$ ⇒$$f=3F5put$$ $$F=80$$ N⇒$$f=48Nfmaximum$$ is $$60$$ N. Therefore, for $$F=80$$ N , both will move together.  Their combined acceleration, by applying NLM using  both as system, $$F=25a$$⇒$$a=8025=3.2$$ $$m/s2$$

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