An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #18 KJ# to # 4KJ# over #t in [0, 9 s]#. What is the average speed of the object?

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Answer 1 · 2017-03-16T20:11:39

The average speed is #=59.38ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The initial velocity is #=u_1#

#1/2m u_1^2=18000J#

The final velocity is #=u_2#

#1/2m u_2^2=4000J#

Therefore,

#u_1^2=2/6*18000=6000m^2s^-2#

and,

#u_2^2=2/6*4000=1333.3m^2s^-2#

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

The points are #(0,6000)# and #(9,1333.3)#

The equation of the line is

#v^2-6000=(1333.3-6000)/9t#

#v^2=-518.5t+6000#

So,

#v=sqrt((-518.5t+6000)#

We need to calculate the average value of #v# over #t in [0,9]#

#(9-0)bar v=int_0^9sqrt(-518.5t+6000) dt#

#9 barv=[(-518.5t+6000)^(3/2)/(3/2*-518.5)]_0^9#

#=((-518.5*9+6000)^(3/2)/(-777.8))-((-518.5*0+6000)^(3/2)/(-777.8))#

#=1333.5^(3/2)/-777.8-6000^(3/2)/-777.8#

#=1/778.5(6000^(3/2)-1333.5^(3/2))#

#=534.4#

So,

#barv=534.4/9=59.38ms^-1#