Question

A body is dropped from a certain height. When it loses $U$ amount of its potential energy it acquires a velocity $v$ . The mass of the body is

• A. $\frac{2U}{v^2}$
• B. $\frac{2v}{U^2}$
• C. $\frac{2v}{U}$
• D. $\frac{U^2}{2v}$

Answer 1

Answer: (A)

Given that
Loss in potential energy, $U_i-U_f=U$
Initial velocity, $u=0$
Final velocity $=v$ .
Let mass of the body is $=m$
Using the conservation of mechanical energy,
$U_i+K_i=U_f+K_f$
$\Rightarrow U_i+0=U_f+\frac{1}{2}m{v}^2$
$\Rightarrow U_i-U_f=\frac{1}{2}m{v}^2$
$\Rightarrow U=\frac{1}{2}m{v}^2$
$\Rightarrow m=\frac{2U}{v^2}$