A uniform chain of length 2,m is kept on a ta | vedclass

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

Question

(B) Let the total length of the chain be $L = 2\,m$ and total mass be $M = 4\,kg$.The length of the hanging part is $l = 60\,cm = 0.6\,m$.The mass per

Vedclass · Accepted Answer

$3.6$

Vedclass · Answer

(B) Let the total length of the chain be $L = 2\,m$ and total mass be $M = 4\,kg$.The length of the hanging part is $l = 60\,cm = 0.6\,m$.The mass per unit length of the chain is $\lambda = \frac{M}{L} = \frac{4}{2} = 2\,kg/m$.The mass of the hanging part is $m = \lambda 	imes l = 2 	imes 0.6 = 1.2\,kg$.The center of mass of the hanging part is at a distance $h = \frac{l}{2} = \frac{0.6}{2} = 0.3\,m$ below the edge of the table.The work done in pulling the chain onto the table is equal to the increase in the potential energy of the hanging part,which is $W = mgh$.Substituting the values: $W = 1.2 	imes 10 	imes 0.3 = 3.6\,J$.

$A$ 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 onto the table? (Take $g = 10\,m/s^2$) ................ $J$

Similar Questions

The potential energy of a particle is given by $U = 8x^2 - 4x + 400 \, J$. Find the position $x$ where the force acting on the particle is zero.

$A$ particle with constant total energy $E$ moves in one dimension in a region where the potential energy is $U(x)$. The speed of the particle is zero where

$A$ chain of mass $M$ and length $L$ is placed on a smooth table such that $1/n$ of its length is hanging over the edge of the table. The hanging part of the chain is pulled back onto the surface of the table. Find the work done to pull it up.

$A$ particle is placed at the origin and a force $F = kx$ is acting on it (where $k$ is a positive constant). If $U(0) = 0$,the graph of $U(x)$ versus $x$ will be (where $U$ is the potential energy function):

The potential energy of a particle of mass $1 \ kg$ moving freely along the $x$-axis is given by $V(x) = (\frac{x^4}{4} - \frac{x^2}{2}) \ J$. If the total mechanical energy of the particle is $2 \ J$,what is its maximum speed in $m/s$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module