An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #66 KJ# to # 225 KJ# over #t in [0, 8 s]#. What is the average speed of the object?

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Answer 1 · 2017-07-13T06:43:18

The average speed is #=217.3ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=6kg#

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=66000J#

The final kinetic energy is #1/2m u_2^2=225000J#

Therefore,

#u_1^2=2/6*66000=22000m^2s^-2#

and,

#u_2^2=2/6*225000=75000m^2s^-2#

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

The points are #(0,22000)# and #(8,75000)#

The equation of the line is

#v^2-22000=(75000-22000)/8t#

#v^2=6625t+22000#

So,

#v=sqrt((6625t+22000)#

We need to calculate the average value of #v# over #t in [0,8]#

#(8-0)bar v=int_0^8sqrt(6625t+22000))dt#

#8 barv=[((6625t+22000)^(3/2)/(3/2*6625)]_0^8#

#=((6625*8+22000)^(3/2)/(9937.5))-((6625*0+22000)^(3/2)/(9937.5))#

#=75000^(3/2)/9937.5-22000^(3/2)/9937.5#

#=1738.5#

So,

#barv=1738.5/8=217.3ms^-1#

The average speed is #=217.3ms^-1#