Question

A body of mass 3 kg is dropped from rest from a height of 1 m. The K.E of the body before touching the ground is:

• A. 150 J
• B. 29.4 N
• C. 29.4 J
• D. 150 N

Answer 1

The correct option is C 29.4 J

Given:
Mass of the body, $m=3kg$
Initial height of the body, $h_i=1m$

Inital potential energy of the body:
$U_i=mg{h}_i$
$U_i=3\times 9.8\times 1=29.4J$

Since the body is dropped from rest, initial kinetic energy is zero.
$K_i=0$

Finally, the body is at the level of ground. So, the height of the body is zero. Thus, the final potential energy of the body:
$U_f=0$

Using conservation of energy, total mechanical energy is conserved.
$U_i+K_i=U_f+K_f$
$29.4+0=0+K_f$
$K_f=29.4J$

In simple words, the potential energy of the body is converted completely into kinetic energy.