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A man of mass 40 kg is standing on a uniform Planck of mass 60 kg lying on a horizontal friction less ice. The man walks from one end to the other end of the Planck. The distance walked by the man relative to ice is (length of Planck=5 m)

\[\begin {array} {1 1} (a)\;2m & \quad (b)\;3m \\ (c)\;5m & \quad  (d)\;4m \end {array}\]

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1 Answer

$d= \large\frac{mL}{m_1+m_2}$
$\quad= \large\frac{40 \times 5 m}{100}$
$\quad=2m$
answered Nov 21, 2013 by meena.p
 

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