An object has a mass of #12 kg#. The object's kinetic energy uniformly changes from #64 KJ# to # 160 KJ# over #t in [0, 5 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-14T07:00:05

The average speed is #=91.79ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=12kg#

The initial velocity is #=u_1#

#1/2m u_1^2=64000J#

The final velocity is #=u_2#

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

Therefore,

#u_1^2=2/12*64000=10666.67m^2s^-2#

and,

#u_2^2=2/12*160000=26666.67m^2s^-2#

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

The points are #(0,10666.67)# and #(5,26666.67)#

The equation of the line is

#v^2-10666.67=(26666.67-10666.67)/5t#

#v^2=3200t+10666.67#

So,

#v=sqrt((3200t+10666.67)#

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

#(5-0)bar v=int_0^5sqrt((3200t+10666.67))dt#

#5 barv=[((3200t+10666.67)^(3/2)/(3/2*3200)]_0^5#

#=((3200*5+10666.67)^(3/2)/(4800))-((3200*0+10666.67)^(3/2)/(4800))#

#=26666.67^(3/2)/4800-10666.67^(3/2)/4800#

#=458.96#

So,

#barv=458.96/5=91.79ms^-1#