An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #480 KJ# to #12 KJ# over #t in [0, 8 s]#. What is the average speed of the object?

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Answer 1 · 2018-02-27T18:03:40

The average speed is #=272.4ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=6kg#

The initial velocity is #=u_1ms^-1#

The final velocity is #=u_2 ms^-1#

The initial kinetic energy is #1/2m u_1^2=480000J#

The final kinetic energy is #1/2m u_2^2=12000J#

Therefore,

#u_1^2=2/6*480000=160000m^2s^-2#

and,

#u_2^2=2/6*12000=4000m^2s^-2#

The graph of #v^2=f(t)# is a straight line

The points are #(0,160000)# and #(8,4000)#

The equation of the line is

#v^2-160000=(4000-160000)/8t#

#v^2=-19500t+160000#

So,

#v=sqrt(-19500t+160000)#

We need to calculate the average value of #v# over #t in [0,8]#

#(8-0)bar v=int_0^8(sqrt(-19500t+160000))dt#

#8 barv= [(-19500t+160000)^(3/2)/(-3/2*19500)] _( 0) ^ (8)#

#=((-19500*8+160000)^(3/2)/(-29250))-((-19500*0+160000)^(3/2)/(-29250))#

#=160000^(3/2)/29250-4000^(3/2)/29250#

#=2179.4#

So,

#barv=2179.4/8=272.4ms^-1#

The average speed is #=272.4ms^-1#