An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #0 KJ# to # 96KJ# over #t in [0, 6 s]#. What is the average speed of the object?

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Answer 1 · 2018-02-01T17:17:25

The average speed is #=9.94ms^-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=0J#

The final kinetic energy is #1/2m u_2^2=96000J#

Therefore,

#u_1^2=2/4*0=0m^2s^-2#

and,

#u_2^2=2/4*96000=48000m^2s^-2#

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

The points are #(0,0)# and #(6,48000)#

The equation of the line is

#v^2=(48000-0)/6t#

#v^2=8000t#

So,

#v=sqrt(8000t)#

We need to calculate the average value of #v# over #t in [0,6]#

#(6-0)bar v=int_0^6(sqrt(8000t))dt#

#6 barv= [(8000t)^(3/2)/(3/2*8000)] _( 0) ^ (6)#

#=((8000*6)^(3/2)/(12000))-((8000*0)^(3/2)/(12000))#

#=8000^(3/2)/12000#

#=59.63#

So,

#barv=58.63/6=9.94ms^-1#

The average speed is #=9.94ms^-1#