An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #48 KJ# to # 15 KJ# over #t in [0, 8 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-15T12:15:43

The average speed is #=101.23ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=6kg#

The initial velocity is #=u_1#

#1/2m u_1^2=48000J#

The final velocity is #=u_2#

#1/2m u_2^2=15000J#

Therefore,

#u_1^2=2/6*48000=16000m^2s^-2#

and,

#u_2^2=2/6*15000=5000m^2s^-2#

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

The points are #(0,16000)# and #(8,5000)#

The equation of the line is

#v^2-16000=(5000-16000)/8t#

#v^2=-1375t+16000#

So,

#v=sqrt((-1375t+16000)#

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

#(8-0)bar v=int_0^8sqrt((-1375t+16000))dt#

#8 barv=[((-1375t+16000)^(3/2)/(-3/2*1375)]_0^8#

#=((-1375*8+16000)^(3/2)/(-2062.5))-((-1375*0+16000)^(3/2)/(-2062.5))#

#=16000^(3/2)/2062.5-5000^(3/2)/2062.5#

#=809.84#

So,

#barv=809.84/8=101.23ms^-1#