Question

(b) A 2 kg object is released from rest from a height of 10 m above the ground. Calculate (i) the initial potential energy at the moment of release; (ii) the kinetic energy at the moment it reaches 4 m above the ground; (iii) the speed of the object just before impact with the ground. After the impact, the object finally comes to rest. Explain what happens to the lost kinetic energy. [Take g = 10 m s?.]
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Answer 1

Answer:

1) 200J

2) 160J

3) 0J

explanation:

Mass = 2kg

Height = 10m

gravity = 10m/s

1) P.E= mgh (mass×gravity×height)

=> 2×10×10= 200J.

2) K.E = mgh (mass×gravity×height)

=> 2×10×4= 160J.

3) K.E= 1/2 ×m×v^2 (m=mass, v=velocity)

=> 1/2 ×2×0= 0J.