A number of identical cubical blocks of edge | vedclass

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

Question

(B) The total number of blocks is $N = n + 1$. Initially, all blocks are on the table, so their center of mass is at height $h_i = a/2$. After stackin

Vedclass · Accepted Answer

$\frac{1}{2}ma\,g\left( {{n^2} + n} \right)$

Vedclass · Answer

(B) The total number of blocks is $N = n + 1$. Initially, all blocks are on the table, so their center of mass is at height $h_i = a/2$. After stacking them into a column of height $H = (n + 1)a$, the center of mass of the column is at $h_f = H/2 = (n + 1)a/2$. The total mass of the system is $M = (n + 1)m$. The work done is equal to the change in gravitational potential energy: $W = \Delta U = M g (h_f - h_i)$. $W = (n + 1)m g \left[ \frac{(n + 1)a}{2} - \frac{a}{2} \right]$. $W = (n + 1)m g \left[ \frac{na + a - a}{2} \right] = (n + 1)m g \left( \frac{na}{2} \right)$. $W = \frac{1}{2} m a g (n^2 + n)$.

$A$ number of identical cubical blocks of edge $a$ and mass $m$ are lying on a horizontal table. The work done on the blocks in arranging them in a column of height $(n + 1)a$ on the table is

Similar Questions

If the potential energy between two molecules is given by $U = -\frac{A}{r^6} + \frac{B}{r^{12}}$,then at equilibrium,the separation between the molecules and the potential energy are:

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

The potential energy of a diatomic molecule is given by $U = \frac{A}{r^{12}} - \frac{B}{r^6}$,where $A$ and $B$ are positive constants. The distance $r$ between them at equilibrium is:

$A$ rod of mass $m$ and length $l$ is lying on a table. What is the work done to make it stand vertically?

$A$ body of weight $1 \, N$ has a potential energy of $1 \, J$ relative to the ground. At what height (in $m$) is it located?

Vedclass Test Series

Exam Paper Generator

Online Exam Module