Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A block of mass $m$ is resting on a smooth horizontal surface. One end of a uniform rope of mass $ \left( \frac{m}{3} \right) $ is fixed to the block, which is pulled in the horizontal direction by applying force $F$ at the other end. The tension in the middle of the rope is

KEAMKEAM 2009Laws of Motion

Solution:

Acceleration of the system
$ a=\frac{F}{\left( m+\frac{m}{3} \right)} $
$ \therefore $ Tension in the middle of the rope $ t=\left( M+\frac{M}{6} \right)a $
Or $ T=\frac{7m}{6}.\frac{3F}{4m} $
Or $ T=\frac{7}{8}F $