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A body of mass 50 kg is pulled by a rope of length 8 m on a surface by a force of 108 N applied at the other end. The force that is acting on the 50 kg mass, if the linear density of rope is 0.5 kg/m will be

\[(a)\;108\; N \quad (b)\;100\; N \quad (c)\;116\; N \quad(d)\;92 \;N \]

1 Answer

mass of rope $=8 \times \large\frac{1}{2}$
$\qquad = 4 \;kg$
Total mass $=54 \;kg$
a-acceleration of system $=\large\frac{F}{m}$
$\qquad=\large\frac{108}{54}$$=2 m/s^2$
Force used for pulling rope $=4 \times 2=8\;N$
Therefore Force applied on mass $=108-8=100 \;N$
Hence b is the correct answer.


answered Jul 5, 2013 by meena.p
edited Jan 26, 2014 by meena.p

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