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In the figure above, two blocks of mass 3$m$ and 2$m$ and 2$m$ are attached together. The plane is frictionless and the pulley is frictionless and massless. The inclined portion of the plane creates an angle $\theta$ with the horizontal floor. What is the acceleration of the block 2$m$ if both blocks are released from rest (gravity $=g ) ?$(A) 2$m g$(B) $\left(\frac{2}{5}\right) g \sin \theta$(C) $\left(\frac{2}{3}\right) g \sin \theta$(D) $\left(\frac{3}{5}\right) g \sin \theta$

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Chapter 12

Practice Test 2

Section 1

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Hi in the given problem here. This is a smooth election lists horizontal surface which is joined with the inclined surface which is inclined at an angle He top with the horizontal. There is a block having a mass to mm and another block kept here. Over the horizontal surface. It's mars is three a.m. There is a frictionless massless bully and a string is connecting these two blogs passing over the pulley like this. No component of weight of this block which is kept on the inclined plane. The wheat is two MG acting vertically downward and one of the component, if this angle is heat at this angle will also be theater. This component will be too MG cost leader and have this component that is to MG sine teacup which will be sliding this block in downward direction. Suppose with an acceleration A. So attention T. Will be created in this spring equal in magnitude in a positive direction about this bully like this. So taking everybody die ground of the smaller block, smaller block which has been kept over the inclined surface, taking it as the block is moving downward. So out of the two forces acting on it. UnG Scientist A and T. These two MG scientists should be more so two MG sine. Theta minus tension is equal to using net force is equal to m into a using Newton's second law of motion. So this is the net force acting on the blog will be kept equal to mass, the product of mass with acceleration and the masses to him. Exhibition support. This is a which will be seen on this block also which is kept on the horizontal surface. This is the question number one. Now, taking the free body diagram of the amir of lock means block having a mass of three um which has been kept kept over the horizontal surface and in the direction of motion there is only one force acting on it which is equal to the tension in the string. So we can say it is only for speedy again. Using Newton's second law of motion, they should be able to three M. Into A. So now putting this value of T. From question to question one, we get two MG sine tita minus three M. A. Is equal to M. A. Or if we shift this three may towards the right hand side, it will become less positive. So two MG sine theta will come out to be equal to five M. A. And cancelling these mars. An expression for the acceleration in the system comes out to be G. Scientists are divided by five which is an answer for this given problem and we can say option B is correct. Thank you.

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