Question

A plate of mass $m$, length $b$ and breadth $a$ is initially lying on a horizontal floor with length parallel to the floor and breadth perpendicular to the floor. Find the work done to erect it on it’s breadth.

• A. $mg(b-a)$
• B. $\frac{mg(b-a)}{2}$
• C. $\frac{mga}{2}$
• D. $\frac{mgb}{2}$

Answer 1

The correct option is B $\frac{mg(b-a)}{2}$

Let $h_1$ and $h_2$ be the position of centre of mass of the plate for two different positions as shown below.



Gravitational potential energy of the plate for position $\text{I}$ and position $\text{II}$ are given by

$U_1=mg{h}_1;U_2=mg{h}_2$

And we know that work done by conservative force is given by

$W_{con}=-\Delta U=-(U_2–U_1)$

$W_{con}=mg{h}_1–mg{h}_2$

where $h_2$ = $\frac{b}{2}$ and $h_1=\frac{a}{2}$

$W_{con}=mg(\frac{a}{2}-\frac{b}{2})=\frac{mg(a-b)}{2}$

From position $\text{I}$ to position $\text{II}$, kinetic energy of plate remains unchanged and equal to zero. So, workdone by external agent will be

$W_e=-W_{con}=\frac{mg(b-a)}{2}$

Hence, option $\left(b\right)$ is the right answer.

Why this question ? Key concept: Work done by conservative force = negative of change in potential energy.