# Application of conservation of mechanical energy for free fall motion – problems and solutions

1. A 1-kg body falls freely from rest, from a height of 80 m. Acceleration due to gravity is 10 m/s^{2}. What is the kinetic energy when the body hits the ground.

__Known :__

Mass (m) = 1 kg

Height (h) = 80 m

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted:__ kinetic energy when the body hits the ground

__Solution :__

The initial mechanical energy (ME_{o}) = gravitational potential energy (PE)

ME_{o} = PE = m g h = (1)(10)(80) = 800 Joule

The final mechanical energy (ME_{t}) = kinetic energy (KE)

The principle of conservation of mechanical energy :

ME_{o} = ME_{t}

PE = KE

800 = KE

The final kinetic energy is 800 Joule.

2. A 4-kg body free fall from rest, from a height of 10 m. Acceleration due to gravity is 10 m s^{–2}. What is the kinetic energy and the velocity at 5 meters above the ground.

__Known :__

The change in height (h) = 10 – 5 = 5 meters

Mass (m) = 4 kg

Acceleration due to gravity (g) = 10 m/s^{2}

Wanted:__ Kinetic energy at 5 meters above the ground and the velocity at 5 meters above the ground__

__Solution :__

__(a) ____Kinetic energy at 5 meters above the ground__

The initial mechanical energy (ME_{o}) = the gravitational potential energy (PE)

ME_{o} = PE = m g h = (4)(10)(5) = 200 Joule

The final mechanical energy (EM_{t}) = kinetic energy (EK)

ME_{t} = KE

The principle of conservation of mechanical energy states that the initial mechanical energy = the final mechanical energy.

ME_{o} = ME_{t}

200 = KE

Kinetic energy at 5 meters above the ground is 200 Joule.

__(b) ____velocity at ____5 meter____s above the ground__

The initial mechanical energy (ME_{o}) = the final mechanical energy (ME_{t})

PE = KE

200 = ½ m v^{2}

2(200) / 4 = v^{2}

100 = v^{2}

v = √100

v = 10 m/s

Body’s velocity at 5 meters above the ground is 10 m/s.

3. A mango falls freely from rest, from a height of 2 meters. Acceleration due to gravity is 10 m s^{–2}. Determine mango’s velocity when hits the ground.

__Known :__

Height (h) = 2 meters

Acceleration due to gravity (g) = 10 m/s^{2}

__Wanted :__ mango’s velocity when hits the ground.

__Solution :__

The initial mechanical energy (ME_{o}) = the gravitational potential energy (PE)

ME = PE = m g h = m (10)(2) = 20 m

The final mechanical energy (ME_{t}) = the kinetic energy (KE)

ME_{t} = KE = ½ m v^{2}

Principle of conservation of mechanical energy states that the initial mechanical energy = the final mechanical energy.

ME_{o} = ME_{t}

20 m = ½ m v^{2}

20 = ½ v^{2}

2(20) = v^{2}

40 = v^{2}

v = √40 = √(4)(10) = 2√10 m/s

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