Minimum Launch Energy

Classical Mechanics Level 5

In the x y z xyz coordinate system, a 1 kg 1\text{ kg} particle is launched from the origin with initial velocity ( v x , v y , v z ) v_x,v_y,v_z) ( m / s ) (m/s) .

The particle experiences a constant acceleration ( a x , a y , a z ) = ( 1 , 5 , 9 ) ( m / s 2 ) (a_x,a_y,a_z) = (1,5,9) (m/s^2) while in motion.

What is the minimum launch energy (in Joules) for the particle to travel through the point ( x , y , z ) = ( 2 , 3 , 7 ) ( m ) (x,y,z) = (2,3,7) (m) ? Give your answer to 2 decimal places.


The answer is 0.72.

Steven Chase by Steven Chase , 37, USA

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1 solution

Prakhar Bindal
Mar 4, 2017

Well the problem is pretty simple

by using equation of kinematics for constant acceleration we get

tvx+t^2/2 = 2

tvy+5t^2/2 = 3

tvz+9t^2/2 = 7

Energy = 1/2 (vx^2 + vy^2+vz^2)

From 3 equations substitute vx,vy,vz into above equation on solving you will get

E = 0.5 (62/t^2 + 107t^2/4 - 80)

Use AM-GM To minimize the t^2 terms

Minimum energy = 0.5(root(6634)-80) = 0.7246 J

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Cool. It was harder my way because I retained more unique variables.

Steven Chase - 4 years, 3 months ago

But where is the potential energy of the conservative force if that is accounted then we get 80.7246

dark angel - 1 year, 1 month ago

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