An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #104 KJ# to # 72KJ# over #t in [0,5s]#. What is the average speed of the object?

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Answer 1 · 2017-03-15T12:45:53

The average speed is #=209.5ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The initial velocity is #=u_1#

#1/2m u_1^2=104000J#

The final velocity is #=u_2#

#1/2m u_2^2=72000J#

Therefore,

#u_1^2=2/4*104000=52000m^2s^-2#

and,

#u_2^2=2/4*72000=36000m^2s^-2#

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

The points are #(0,52000)# and #(5,36000)#

The equation of the line is

#v^2-52000=(36000-52000)/5t#

#v^2=-3200t+52000#

So,

#v=sqrt((-3200t+52000)#

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

#(5-0)bar v=int_0^5sqrt(-3200t+52000) dt#

#5 barv=[(-3200t+52000)^(3/2)/(3/2*-3200)]_0^5#

#=((-3200*5+52000)^(3/2)/(-48000))-((-3200*0+52000)^(3/2)/(-48000))#

#=36000^(3/2)/-4800-52000^(3/2)/-4800#

#=1/4800(52000^(3/2)-36000^(3/2))#

#=1047.4#

So,

#barv=1047.4/5=209.5ms^-1#