An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #128 KJ# to # 48 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

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Answer 1 · 2017-06-01T18:32:42

The average speed is #=207.9ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=128000J#

The final velocity is #=u_2#

#1/2m u_2^2=48000J#

Therefore,

#u_1^2=2/4*128000=64000m^2s^-2#

and,

#u_2^2=2/4*48000=24000m^2s^-2#

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

The points are #(0,64000)# and #(12,24000)#

The equation of the line is

#v^2-64000=(24000-64000)/12t#

#v^2=-3333.3t+64000#

So,

#v=sqrt((-3333.3t+6400)#

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

#(12-0)bar v=int_0^12sqrt((-3333.3t+64000))dt#

#12 barv=[((-3333.3t+64000)^(3/2)/(3/2*3333.3)]_0^12#

#=((-3333.3*12+64000)^(3/2)/(5000))-((-3333.3*0+64000)^(3/2)/(5000))#

#=64000^(3/2)/5000-24000^(3/2)/5000#

#=2494.6#

So,

#barv=2494.6/12=207.9ms^-1#