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In the arrangement shown pullies and strings are ideal. The acceleration of block B is

Asked by soulful.my.me | 11 Aug, 2023, 05:27: PM

Expert Answer

Figure shows the free body diagram of block-B , block-A and of pulley.

In Block-B , (mg) is weight of the block and T is tension force in the rope . Le block-B moves upward with acceleration a.

In Pulley , Tension force in the cable connected to block A becomes 2T because there are two tension forces acting in the

direction opposite to tension force of the cable connected to block-A .

In block-A , (2mg) is weight and N is normal force .

If there is no displacement in vertical direction , then net force along vertical direction is zero.

Hence we get , N - ( 2mg ) = 0 , or normal force N = ( 2mg )

Friction force f = μN , where μ = 0.5 is kinetic friction coefficient .

Hence friction force f = 0.5 * ( 2mg ) = mg

If we apply Newton's second law to the block-B , then we have

begin mathsize 14px style T space minus space m g space equals space m space a end style

............................(1)

If we apply Newton's second law to the block-A , then we have

begin mathsize 14px style left parenthesis 4 m g right parenthesis space minus space 2 T space minus space f space equals space left parenthesis 2 m right parenthesis space left parenthesis a divided by 2 right parenthesis end style

By substituting friction force and after simplification , above expression becomes

begin mathsize 14px style left parenthesis 4 m g right parenthesis space minus space 2 T space minus space left parenthesis m g right parenthesis space equals space m space a end style

......................(2)

We can eliminate tesnion force T by multiplying eqn.(1) by 2 and addig to eqn.(2) ,

then we get after simplification

begin mathsize 14px style g space equals space 3 space a end style

hence acceleration of block-B is , a = (1/3) g

Answered by Thiyagarajan K | 12 Aug, 2023, 11:40: AM

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