question_answer

DIRECTIONS (Qs 51): Read the following passage and answer the questions that follows

A block of mass \[15\text{ }kg\] is placed over a frictionless horizontal surface. Another block of mass \[10\text{ }kg\] is placed over it, that is connected with a light string passing over two pulleys fastened to the \[15\text{ }kg\] block. A force \[F=80N\]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 \[g=10\text{ }m/{{s}^{2}}\])

The magnitude of acceleration of the \[10\text{ }kg\] block is

A)  \[3.2\text{ }m/{{s}^{2}}\]                  

B)  \[2.0\text{ }m/{{s}^{2}}\]

C)  \[1.6\text{ }m/{{s}^{2}}\]                  

D)  \[0.8\text{ }m/{{s}^{2}}\]

Correct Answer: A

Solution :

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 \[10\text{ }kg\] block will not slip over the \[15\,kg\] block till acceleration of \[15\,kg\] block becomes maximum as it is created only by friction force exerted by \[10\,kg\] block on it. \[{{a}_{1}}>{{a}_{2(\max )}}\] \[\frac{F-f}{10}=\frac{f}{15}\] for limiting condition as f maximum is 60N. \[F=100N\] Therefore, for \[F=80\,N,\] both will move together. Their combined acceleration, by applying NLM using both as system, \[F=25a\]                 \[a=\frac{80}{25}=3.2\,m/{{s}^{2}}\]