An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #15 KJ# to #135 KJ# over #t in [0, 9 s]#. What is the average speed of the object?

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Answer 1 · 2017-06-10T15:16:50

The average speed is #=167.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=5kg#

The initial velocity is #=u_1#

The initial kinetic energy is #1/2m u_1^2=15000J#

The final velocity is #=u_2#

The final kinetic energy is #1/2m u_2^2=135000J#

Therefore,

#u_1^2=2/5*15000=6000m^2s^-2#

and,

#u_2^2=2/5*135000=54000m^2s^-2#

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

The points are #(0,6000)# and #(9,54000)#

The equation of the line is

#v^2-6000=(54000-6000)/9t#

#v^2=5333.3t+6000#

So,

#v=sqrt((5333.3t+6000)#

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

#(9-0)bar v=int_0^9sqrt((5333.3t+6000))dt#

#9 barv=[((5333.3t+6000)^(3/2)/(3/2*5333.3)]_0^9#

#=((5333.3*9+6000)^(3/2)/(8000))-((5333.3*0+6000)^(3/2)/(8000))#

#=54000^(3/2)/8000-6000^(3/2)/8000#

#=1510.5#

So,

#barv=1510.5/9=167.8ms^-1#

The average speed is #=167.8ms^-1#