An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #84 KJ# to # 48 KJ# over #t in [0, 6 s]#. What is the average speed of the object?

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Answer 1 · 2018-02-21T10:09:43

The average speed is #=147.9ms^-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=84000J#

The final kinetic energy is #1/2m u_2^2=48000J#

Therefore,

#u_1^2=2/6*84000=28000m^2s^-2#

and,

#u_2^2=2/6*48000=16000m^2s^-2#

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

The points are #(0,28000)# and #(6,16000)#

The equation of the line is

#v^2-28000=(16000-28000)/6t#

#v^2=-2000t+28000#

So,

#v=sqrt(-2000t+28000)#

We need to calculate the average value of #v# over #t in [0,6]#

#(6-0)bar v=int_0^6(sqrt(-2000t+28000))dt#

#6 barv= [(-2000t+28000)^(3/2)/(-3/2*2000)] _( 0) ^ (6)#

#=((-2000*6+28000)^(3/2)/(-3000))-((-2000*0+28000)^(3/2)/(-3000))#

#=28000^(3/2)/3000-16000^(3/2)/3000#

#=887.15#

So,

#barv=887.15/6=147.9ms^-1#

The average speed is #=147.9ms^-1#