Question

a body of mass 5 kg Falls from the rest from a height of 10 m from the Earth surface the kinetic energy of the body on the earth surface will be

Answer 1

mgh=1/2mv^2
[2gh]^1/2=v
(2×5×10)^1/2
10

10m/s is a speed

Answer 2

Answer: The correct answer is 490 J.

Explanation:

It is given in the problem that a body of mass 5 kg Falls from the rest from a height of 10 m from the Earth surface.

From the energy conservation law,

P.E= Final kinetic energy - initial kinetic energy

$mgh= \frac{1}{2} mv^{2} -\frac{1}{2} mu^{2}$

Here, m is the mass of the object, h is the height, v is the final velocity and u is the initial kinetic energy.

Put h= 10 m, g=9.8 meter per second square, m= 5 kg and u= 0.

$(5)(9.8)(10)= \frac{1}{2}mv^{2} -\frac{1}{2}m(0)^{2}$

$\frac{1}{2}mv^{2}=490 J$

Therefore, the kinetic energy of the body on the earth surface is 490 J.