Question

a body of mass 5 kg is dropped from a height of 1m its kinetic energy just before reaching the ground is​

Answer 1

SOLUTION

As we know when a body dropped for a height h then the velocity of the body goes on increasing because the direction of acceleration due to gravity and the direction of the velocity vector is same.

Initial velocity = 0

Final velocity = v

Acceleration = g

Height = 1m

ACC to EQ of motion

v² = u² + 2gh

v² = 0² + 20h

v² = 20m/sec

Kinetic energy of the body with velocity v = 1/2mv²

Kinetic energy = 1/2×5×20

Kinetic energy = 50Joules

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Answer 2

Solution:

Given:

➜ A body of mass 5 kg is dropped from a height of 1m.

Find:

➜ Find what is kinetic energy just before reaching the ground.

According to the given question:

➜ Mass = 5 kg

➜ Height = 1 m

Know terms:

➜ Initial velocity = (u)

➜ Final velocity = (v)

➜ Acceleration = (a)

➜ Joules = (j)

Calculations:

Accordingly to equation of motion:

➜ v² = u² + 2 gh

➜ v² = 0² + 20 h

➜ v² = 20 m/sec

v = 1/2 mv² => kinetic energy of the body with velocity.

➜ Kinetic energy = 1/2 × 5 × 20

➜ Kinetic energy = 50 J

Therefore, 50 joules its kinetic energy just before reaching the ground.

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