An object has a mass of #8 kg#. The object's kinetic energy uniformly changes from #640 KJ# to # 320 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

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Answer 1 · 2017-10-24T05:47:54

The average speed is #=344.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=8kg#

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=640000J#

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

Therefore,

#u_1^2=2/8*640000=160000m^2s^-2#

and,

#u_2^2=2/8*320000=80000m^2s^-2#

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

The points are #(0,160000)# and #(12,80000)#

The equation of the line is

#v^2-160000=(80000-160000)/12t#

#v^2=-6666.7t+160000#

So,

#v=sqrt((-6666.7t+160000)#

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

#(12-0)bar v=int_0^12(sqrt(-6666.7t+160000))dt#

#12 barv=[((-6666.7t+160000)^(3/2)/(-3/2*6666.7))]_0^12#

#=((-6666.7*12+160000)^(3/2)/(-10000))-((-1666.7*0+160000)^(3/2)/(-10000))#

#=160000^(3/2)/10000-80000^(3/2)/10000#

#=4137.3#

So,

#barv=4137.3/12=344.8ms^-1#

The average speed is #=344.8ms^-1#