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  4. The velocity of an 800 g object changes from v0 =(3i-4j)ms-1 to vf =(6j+2K)ms-1change in kinetic energy is

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Q. The velocity of an 800 g object changes from $v_0 =(3i-4j)ms^{-1} \,to \,v_f =(6j+2K)ms^{-1}$change in kinetic energy is

KEAMKEAM 2004Work, Energy and Power

Solution:

Given m = 800 g = 0.8 kg
$v_0=(3i-4j)ms^{-1}$
$v_0=v_0\,v_0=(3i-4j).(3i-4j)$
= 9+16 = 25
$v_f=(-6j+2k)ms^{-1}$
$v^2_f=v_f.v_f=(-6j+2k)(-6j+2k)$
$=36 +4=40 $
Change in KE =$\frac{1}{2}m(v^2_f-v^2_0)$
$\frac{1}{2}\times 0.8 (40 -25)=6 J$