# NEET Class neet Answered

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Asked by jhajuhi19 | 23 Nov, 2019, 02:12: PM

Expert Answer

Figure given above shows the free body diagram (FBD) of bith blocks. Forces acting on each block are shown in FBD .

Let us consider the block of mass M. Mg is the weight , i.e. gravitational force of block,

N

_{2}is the reaction force acting beteen the contact surfaces of floor and block,μ

_{2}N_{2}is the friction force acting against the movement of block .T is tension force acting on the rope and aslo at pulley connected to block.

N

_{1}is the normal reaction force exerted by block of mass mNow let us see FBD of block of mass m . Gravitationa forec, i.e. weight mg is acting downward, T is tension in the rope,

μ

_{1}N_{1}is the friction force acting against the movement along the contact surface between two blocksand N

_{1}is the reaction force exerted by block of mass M.Let us apply the Newtons second law for the block of mass m

In horizontal direction, if we assume a

_{x}is acceleration of mass m , then we have, N_{1}= m a_{x}.....................(1)In vertical direction, if we assume a

_{y}is acceletaion of mass m , then we have, mg - μ_{1}N_{1}- T = m a_{y}...............(2)Let us apply the Newtons second law for the block of mass M

In horizontal direction, if we assume A

_{x }is_{ }the acceleration of block of mass M,then we have, 2T - N

_{1}- μ_{2}N_{2}= M A_{x}...............(3)In Vertical direction, there is no movement , hence we have, N

_{2}= M g + T ........................(4)At equilibrium, we have, a

_{x}= a_{y}= A_{x}= 0Hence from eqn.(1), we get N

_{1}= 0By using N

_{1}= 0 and a_{y }= 0, from eqn.(2) we get, T = mgBy substituting for T, N

_{1 }, N_{2}and A_{x}in eqn.(3) , we get , ( 2 m g ) - μ_{2}( M g + m g ) = 0Hence , we get, m = ( μ

_{2}M ) / ( 2 - μ_{2}) = (0.4 × 20 ) / (2 - 0.4 ) = 5 kg
Answered by Thiyagarajan K | 23 Nov, 2019, 10:55: PM

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