Conservation of Mechanical Energy  – Falling body without air resistance

Example

From a height of 15 m an object with mass of 2.0 kg  is thrown down with an initial velocity of  3.0 ms-1 .

Find the velocity at which the object the ground reaches.

The mechanical energy is constant, because there is no friction.

Ek + Eg = constant.  

The heighst point is called H and lowest point G

(Ek + Eg) H  =( Ek + Eg) G

  ½(2.0) (3.0)2 + (2.0)( 9.81)( 15) = Ek + 0        (h = 0 m)

9.0 + 294.3 = Ek,G  

E k,G = 303.3 J

E k,G = ½  (2.0) (v)2

303.3 = ½ (2.0)(v)2

v = 17.4 m/s

Remark

Because the mass is a factor in every term , the mass can be eliminated.

We can write:

   ½ (vH)2 + g hH = ½ (vG)2

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