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

Question

A body is dropped from a certain height above the surface of the earth. When it loses U amount of its potential energy, it acquires a velocity v . The mass of the body is (A) (2 U/v2) (B) (2 v/U2) (C) (2 v/U) (D) (U2/2 v)

Tardigrade Admin · Accepted Answer

Using the conservation of mechanical energy Ki+Pi=Kf+Pf 0+P=(1/2)mv2+(P - U) (1/2)mv2=U m=(2 U/v2)

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

$\frac{2 U}{v ^{2}}$

$\frac{2 v}{U ^{2}}$

$\frac{2 v}{U}$

$\frac{U ^{2}}{2 v}$

Using the conservation of mechanical energy
$K_{i} + P_{i} = K_{f} + P_{f}$
$0 + P = \frac{1}{2} m v^{2} + (P - U)$
$\frac{1}{2} m v^{2} = U$
$m = \frac{2 U}{v ^{2}}$

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

v22U​

U22v​

U2v​

2vU2​