Question

A body of mass $5\text{kg}$ is dropped from a tower of height $10\text{m}$. Find the difference between work done $\text{(}J)$ by gravity and change in potential energy of the body from ${}^{\prime }{A}^{\prime }$ to ${}^{\prime }{B}^{\prime }$ (Take $g=10\frac{m}{s^2}$)

• A. $-1000\text{joule}$
• B. $-100\text{joule}$
• C. $1000\text{joule}$
  
• D. $2000\text{joule}$

Answer 1

The correct option is C $1000\text{joule}$


Work done by gravity from '$A$' to '$B$'
$W_{AB}=+mgh$;
because gravity is acting downwards and body is also getting displaced downwards.

$\Rightarrow$Now change in potential energy from '$A$' to '$B$'
$\Delta U_{A\to B}=U_B-U_A...\left(i\right)=mg{h}_B-mg{h}_A=mg\left(0\right)-mgh=-mgh$
(Where $B$ is the refrence point)

Now, difference between workdone ($\text{J}$) by gravity and change in potential energy of body from ${}^{\prime }{A}^{\prime }$ to ${}^{\prime }{B}^{\prime }$ is given by
$W_{A\to B}-U_{A\to B}=(+mgh-(-mgh\left)\right)=2mgh=2\times 5\times 10\times 10=1000\text{joule}$