Question

An object of mass 4kg is dropped from a tower of 50 m . Find the kinetic energy possessed by the object when it is just about to reach the bottom tower

Answer 1

GIVEN:- Mass of the object = 4 kg

Height of the tower = 50 m.

To find kinetic energy at the

just before bottom of the

tower.

ANSWEN:-

By using law of conservation of energy.

(initial potential energy + initial kinetic energy = final potential energy + final kinetic energy )

{ mgh + (1/2)mv² = mgh + (1/2) mv² }

{ 4×10×50 + (1/2)×4×(0)² = 4×10×0 + (1/2)×4×(10√10)² }

{ 2000 + 0 = 0 + 2000)

[where v = 10√10, by using v²= u² + 2gh ]

OR kinetic energy is obtain by the formula ,

K.E. = (1/2)mv²

= (1/2)×4×(10√10)²

= (1/2)×4×100×10

= (1/2)×4×1000

= (1/2)×4000

= 2000 Joule

Therefore, the kinetic energy of the object just before bottom of tower is 2000 joule.

THANKS.