# Application of conservation of mechanical energy for motion on inclined plane – problems and solutions

1. A block slides down on smooth inclined plane without friction. What is block’s velocity when hits the ground. Acceleration due to gravity is 10 m/s^{2}

__Known :__

Height (h) = 8 m

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

__Wanted__ : velocity (v)

__Solution__ :

__Initial mechanical energy (ME___{o}__) = gravitational potential energy (PE)__

ME_{o} = PE = m g h = m (10)(8) = 80 m

__Final mechanical energy (ME___{t}__) = 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}

80 m = ½ m v^{2}

80 = ½ v^{2}

160 = v^{2}

v = √160 = √(16)(10) = 4√10 m/s

2. A 1-kg object slides down along 8 meters. Determine kinetic energy after the object moves along 5 meters… Acceleration due to gravity g = 10 m/s^{2}

__Known :__

Mass (m) = 0.2 kg

d = 5 meters

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

__Wanted__ : kinetic energy (KE)

__Solution :__

sin 30^{o} = h / d

0.5 = h / 5

h = (0.5)(5) = 2.5 meters

The change in height of the object is 2.5 meters.

__The initial mechanical energy (ME___{o}__) = the gravitational potential energy (PE)__

ME_{o} = PE = m g h = (1)(10)(2.5) = 25 Joule

__The final mechanical energy (ME___{t}__) = kinetic energy (KE)__

ME_{t} = KE

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

ME_{o} = ME_{t}

25 = KE

Kinetic energy = 25 Joule.

**Ebook PDF Application of conservation of mechanical energy for motion on inclined plane 43.06 KB**

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