A body of mass 10 kg is dropped from the top of a tower. The height of tower is 10 m. Calculate:
Initial potential energy of the body
Its potential energy at 5 m above the ground
Maximum value of its kinetic energy.
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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.
Explanation:
hope it helps
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