An object has a mass of #8 kg#. The object's kinetic energy uniformly changes from #24 KJ# to # 42KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-06-14T05:50:20

The average speed is #=90.54ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=8kg#

The initial velocity is #=u_1#

The initial kinetic energy is #1/2m u_1^2=24000J#

The final velocity is #=u_2#

The final kinetic energy is #1/2m u_2^2=42000J#

Therefore,

#u_1^2=2/8*24000=6000m^2s^-2#

and,

#u_2^2=2/8*42000=10500m^2s^-2#

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

The points are #(0,6000)# and #(3,10500)#

The equation of the line is

#v^2-6000=(10500-6000)/3t#

#v^2=1500t+6000#

So,

#v=sqrt((1500t+6000)#

We need to calculate the average value of #v# over #t in [0,3]#

#(3-0)bar v=int_0^3sqrt((1500t+6000))dt#

#3 barv=[((1500t+6000)^(3/2)/(3/2*1500)]_0^3#

#=((1500*3+6000)^(3/2)/(2250))-((1500*0+6000)^(3/2)/(2250))#

#=10500^(3/2)/2250-6000^(3/2)/2250#

#=271.63#

So,

#barv=271.63/3=90.54ms^-1#

The average speed is #=90.54ms^-1#