An object has a mass of #9 kg#. The object's kinetic energy uniformly changes from #81 KJ# to # 18 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

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Answer 1 · 2017-08-30T06:44:12

The average speed is #=103.0ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=9kg#

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=81000J#

The final kinetic energy is #1/2m u_2^2=18000J#

Therefore,

#u_1^2=2/9*81000=18000m^2s^-2#

and,

#u_2^2=2/9*18000=4000m^2s^-2#

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

The points are #(0,18000)# and #(4,4000)#

The equation of the line is

#v^2-18000=(4000-18000)/4t#

#v^2=-3500t+18000#

So,

#v=sqrt((-3500t+18000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4(sqrt(-3500t+18000))dt#

#4 barv=[((-3500t+18000)^(3/2)/(-3/2*3500))]_0^4#

#=((-3500*4+18000)^(3/2)/(-5250))-((-3500*0+18000)^(3/2)/(-5250))#

#=18000^(3/2)/5250-4000^(3/2)/5250#

#=411.8#

So,

#barv=411.8/4=103.0ms^-1#

The average speed is #=103.0ms^-1#