Question
Question: A block of mass M is pulled along a horizontal frictionless surface by a rope of mass \[{M {\left/ ...
A block of mass M is pulled along a horizontal frictionless surface by a rope of mass M/M22. If a force 2Mg is applied at one end of the rope, the force which the rope exerts on the block is –
Solution
Based on the concept of Newton’s second law of motion, we can say that force exerted by the rope on the block is equal to the product of acceleration and total mass of rope and block system.
Complete step by step answer:
Given:
The mass of the block is M.
The mass of rope is M/M22.
Force applied on one end of the rope is P=2Mg.
The block will start moving under the action of force P with acceleration, and the value of that acceleration is given by:
a=totalmasstotalexternalforce……(1)
We know that the value of total external force is given by force P, and the total mass is equal to the summation of the mass of the block and rope.
Total mass $$= M + {M {\left/
{\vphantom {M 2}} \right.
} 2}\\
\Rightarrow\dfrac{{3M}}{2}
a = \dfrac{{2Mg}}{{\dfrac{{3M}}{2}}}\\
\Rightarrow a = \dfrac{{4g}}{3}
F = \dfrac{{3M}}{2} \times \dfrac{{4g}}{3}\\
\therefore F = 2Mg