1.

A 10 kg ball is dropped from a height of 10m. Find(a) the initial potential energy of the ball , (b) the kinetic energy just before it reaches the ground, and (c) the speed just before it reaches the ground. (Ans: 980J, 980J, 14m/s)

Answer»

(a). Mass of the Body = 10 kg.

Height = 10 m.

Acceleration DUE to gravity = 9.8 m/s².


Using the Formula,

Potential Energy = mgh

= 10 × 9.8 × 10

= 980 J.


(B). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.


∴ Kinetic Energy before the body REACHES the ground is equal to the Potential Energy at the height of 10 m.


∴ Kinetic Energy = 980 J.



(c). Kinetic Energy = 980 J.

Mass of the ball = 10 kg.


∵ K.E. = 1/2 × mv²

∴ 980 = 1/2 × 10 × v²

∴ v² = 980/5

⇒ v² = 196

∴ v = 14 m/s.



Hope it helps.



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