A 5 kg of ball is chopped from a height of 10 m (1) Find the potential energy of the ball.(2)Find the kinetic energy of the ball.(3)Calculate the velocity before reaches the ground.
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Mass of the Body = 5 kg.
Height = 10 m.
Acceleration due to gravity = 9.8 m/s².
Using the Formula, Potential Energy = mgh
= 5 × 9.8 × 10 = 490 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 = 490 J.
(c). Kinetic Energy = 490 J.
Mass of the ball = 5 kg.
∵ K.E. = 1/2 × mv²
∴ 490 = 1/2 × 5 × v²
∴ v² = 490 × 2/5
⇒ v² = 196
∴ v = 14 m/s.
Sorry i take mass as 10 kg. I edited my answer again.
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