Question 229: Laws of Motion

Question

A block A of mass m
1
rests on a horizontal table. A light string connected to it passes over a frictionless pully at the edge of table and from its other end another block B of mass m
2
is suspended. The coefficient of kinetic friction between the block and the table is $$\mu $$
k
.  When the block A is sliding on the table,  the tension in the string is

Accepted Answer

For the motion of both the blocks
m
1
a = T Ã¢ÂÂ $${\mu _k}$$m
1
g
m
2
g Ã¢ÂÂ T = m
2
a
$$a = {{{m_2}g - {\mu _k}{m_1}g} \over {{m_1} + {m_2}}}$$
$${m_2}g - T = \left( {{m_2}} \right)\left( {{{{m_2}g - {\mu _k}{m_1}g} \over {{m_1} + {m_2}}}} \right)$$
solving we get tension in the string
$$T = {{{m_1}{m_2}\left( {1 + {\mu _k}} \right)g} \over {{m_1} + {m_2}}}$$

Answer

$${{{m_1}{m_2}(1 + {\mu _k})g} \over {({m_1} + {m_2})}}$$

Answer

$${{{m_1}{m_2}(1 - {\mu _k})g} \over {({m_1} + {m_2})}}$$

Answer

$${{\left( {{m_2} + {\mu _k}{m_1}} \right)g} \over {\left( {{m_1} + {m_2}} \right)}}$$

Answer

$${{\left( {{m_2} - {\mu _k}{m_1}} \right)g} \over {\left( {{m_1} + {m_2}} \right)}}$$

Laws of Motion

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