An object has a mass of #12 kg#. The object's kinetic energy uniformly changes from #96 KJ# to # 160 KJ# over #t in [0, 5 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-22T05:42:10

The average speed is #=145.7ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=12kg#

The initial velocity is #=u_1#

#1/2m u_1^2=96000J#

The final velocity is #=u_2#

#1/2m u_2^2=160000J#

Therefore,

#u_1^2=2/12*96000=16000m^2s^-2#

and,

#u_2^2=2/12*160000=26666.7m^2s^-2#

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

The points are #(0,16000)# and #(5,26666.7)#

The equation of the line is

#v^2-16000=(26666.7-16000)/5t#

#v^2=2133.3t+16000#

So,

#v=sqrt((2133.3t+16000)#

We need to calculate the average value of #v# over #t in [0,5]#

#(5-0)bar v=int_0^5sqrt((2133.3t+16000))dt#

#5 barv=[((2133.3t+16000)^(3/2)/(3/2*2133.3)]_0^5#

#=((2133.3*5+16000)^(3/2)/(3200))-((213303*0+16000)^(3/2)/(3200))#

#=26666.7^(3/2)/3200-16000^(3/2)/3200#

#=728.37#

So,

#barv=728.37/5=145.7ms^-1#