An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #96 KJ# to # 320 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

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Answer 1 · 2018-03-18T15:52:54

The average speed is #=318.4ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=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=96000J#

The final kinetic energy is #1/2m u_2^2=320000J#

Therefore,

#u_1^2=2/4*96000=48000m^2s^-2#

and,

#u_2^2=2/4*320000=160000m^2s^-2#

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

The points are #(0,48000)# and #(12,160000)#

The equation of the line is

#v^2-48000=(160000-48000)/12t#

#v^2=9333.3t+48000#

So,

#v=sqrt(9333.3t+48000)#

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

#(12-0)bar v=int_0^12(sqrt(9333.3t+48000))dt#

#12 barv= [(9333.3t+48000)^(3/2)/(3/2*9333.3)] _( 0) ^ (12)#

#=((9333.3*12+48000)^(3/2)/(14000))-((9333.3*0+48000)^(3/2)/(14000))#

#=16000^(3/2)/14000-48000^(3/2)/14000#

#=3820.3#

So,

#barv=3820.3/12=318.4ms^-1#

The average speed is #=318.4ms^-1#