A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is µk. When the block A is sliding on the table, the tension in the string is
i am getting an ans:g(m1m2(uk+1)+ukm2^2) please tell me hoe to get the orginal ans thank you
ans
m1m2(l+µk )g/(m1 +m2)
Answers
Answered by
220
oh it's okay , Let's try to solve !
question said , a block 'A' of mass m1 rest on horizontal table and a light string connected it over frictionless pulley at the edge of the table from it other end another block of mass m2 is suspended.
when block is sliding , kinetic friction acts on contacting surface of block A and surface of the table as shown in figure.
so, we can write equation ,
.......(i)
similarly, for block B we can write equation,
.......(ii)
from equation (i) and (ii), we get,
now put a in equation (ii),
question said , a block 'A' of mass m1 rest on horizontal table and a light string connected it over frictionless pulley at the edge of the table from it other end another block of mass m2 is suspended.
when block is sliding , kinetic friction acts on contacting surface of block A and surface of the table as shown in figure.
so, we can write equation ,
.......(i)
similarly, for block B we can write equation,
.......(ii)
from equation (i) and (ii), we get,
now put a in equation (ii),
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Answered by
97
Hope it helps! Use free body diagram for both masses m1 and m2. Derive equations from them and solve as given.
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