Question

A body of mass 5 kg falls from rest from a height 10 m from
earth's surface. The kinetic energy of the body on the earth's
surface will be (if g 9.8 m/sec2)

Answer 1

The body falls from rest from a height 10m from earth's surface.

so, initial velocity of the body, u = 0 m/s

displacement covered by body to reach the earth's surface, S = 10m

using formula, v² = u² + 2as

here, a = g = 9.8m/s²

now, v² = 0 + 2 × 9.8 × 10 = 196 m²/s²

now kinetic energy of the body on the earth's surface = 1/2 m v²

= 1/2 × 5kg × (196 m²/s²)

= 490 kgm²/s² or 490J

hence, kinetic of the body on the earth's surface will be 490 J

Answer 2

Answer:

The kinetic energy of the body on the surface of Earth is 490J

Explanation:

Given :

mass=m=5kg

Initial velocity =u=0 m/s

Height=h=10m

K.E=?

g= 9.8 m/s²

From Third equation of motion :

v² -u²=2as

where a= g and s=h for freely falling body

v²-u²=2gh

v²-0²=2gh

v=√2gh

=√2x9.8x 10

=√196

=14 m/s

Now kinetic energy :

K.E=1/2 m V²

=1/2 x 5x14x14

= 196x5/2

=490 J

∴The kinetic energy of the body on the surface of Earth is 490J