An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #16 KJ# to #96 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

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Answer 1 · 2017-04-28T06:21:39

The average speed is #=231ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=2kg#

The initial velocity is #=u_1#

#1/2m u_1^2=16000J#

The final velocity is #=u_2#

#1/2m u_2^2=96000J#

Therefore,

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

and,

#u_2^2=2/2*96000=96000m^2s^-2#

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

The points are #(0,16000)# and #(4,96000)#

The equation of the line is

#v^2-16000=(96000-16000)/4t#

#v^2=20000t+16000#

So,

#v=sqrt((20000t+16000)#

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

#(4-0)bar v=int_0^4sqrt((20000t+16000))dt#

#4 barv=[((20000t+16000)^(3/2)/(3/2*20000)]_0^4#

#=((20000*4+16000)^(3/2)/(30000))-((20000*0+16000)^(3/2)/(30000))#

#=96000^(3/2)/30000-16000^(3/2)/30000#

#=924#

So,

#barv=924/4=231ms^-1#