An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #12 KJ# to # 360 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

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Answer 1 · 2017-08-17T06:19:42

The average speed is #=290.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=4kg#

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=12000J#

The final kinetic energy is #1/2m u_2^2=360000J#

Therefore,

#u_1^2=2/4*12000=6000m^2s^-2#

and,

#u_2^2=2/4*360000=180000m^2s^-2#

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

The points are #(0,6000)# and #(12,180000)#

The equation of the line is

#v^2-6000=(180000-6000)/12t#

#v^2=14500t+6000#

So,

#v=sqrt((14500t+6000)#

We need to calculate the average value of #v# over #t in [0,12]#

#(12-0)bar v=int_0^12(sqrt(14500t+6000))dt#

#12 barv=[((14500t+6000)^(3/2)/(3/2*14500))]_0^12#

#=((14500*12+6000)^(3/2)/(21750))-((14500*0+6000)^(3/2)/(21750))#

#=180000^(3/2)/21750-6000^(3/2)/21750#

#=3489.8#

So,

#barv=3489.8/12=290.8ms^-1#

The average speed is #=290.8ms^-1#