A uniform chain has a mass m and length l. It | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The work done in pulling the hanging part of the chain onto the table is equal to the increase in the gravitational potential energy of that part.

Vedclass · Accepted Answer

$\frac{m g l}{72}$

Vedclass · Answer

(A) The work done in pulling the hanging part of the chain onto the table is equal to the increase in the gravitational potential energy of that part.Let the mass of the hanging part be $m' = \frac{m}{6}$ and its length be $l' = \frac{l}{6}$.The center of mass of the hanging part is at a distance $h = \frac{l'}{2} = \frac{l/6}{2} = \frac{l}{12}$ below the edge of the table.The work done $W$ is equal to the potential energy required to raise this center of mass to the level of the table:$W = m' g h$$W = \left(\frac{m}{6}\right) g \left(\frac{l}{12}\right)$$W = \frac{m g l}{72}$

$A$ uniform chain has a mass $m$ and length $l$. It is held on a frictionless table with one-sixth of its length hanging over the edge. The work done in just pulling the hanging part back on the table is:

Similar Questions

$A$ uniform chain of length $2\,m$ is kept on a table such that a length of $60\,cm$ hangs freely from the edge of the table. The total mass of the chain is $4\,kg$. What is the work done in pulling the entire chain on the table? (Take $g = 10\,m/s^2$)

If a stone is thrown vertically upward and returns to the ground,its potential energy is maximum:

Write the equation of total mechanical energy of a body having mass $m$ and stationary at height $H$.

$A$ uniform chain of length $L$ and mass $M$ is lying on a smooth table and one-third of its length is hanging vertically down over the edge of the table. If $g$ is the acceleration due to gravity, the work required to pull the hanging part onto the table is

Potential energy as a function of $r$ is given by $U = \frac{A}{r^{10}} - \frac{B}{r^{5}}$,where $r$ is the interatomic distance,and $A$ and $B$ are positive constants. The equilibrium distance between the two atoms will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module