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