An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #96 KJ# to # 108 KJ# over #t in [0, 6 s]#. What is the average speed of the object?

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Answer 1 · 2017-03-17T09:11:50

The average speed is #=184.4ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The initial velocity is #=u_1#

#1/2m u_1^2=96000J#

The final velocity is #=u_2#

#1/2m u_2^2=108000J#

Therefore,

#u_1^2=2/6*96000=32000m^2s^-2#

and,

#u_2^2=2/6*108000=36000m^2s^-2#

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

The points are #(0,32000)# and #(6,36000)#

The equation of the line is

#v^2-32000=(36000-32000)/6t#

#v^2=666.7t+32000#

So,

#v=sqrt((666.7t+32000)#

We need to calculate the average value of #v# over #t in [0,6]#

#(6-0)bar v=int_0^6sqrt((666.7t+32000)dt#

#6 barv=[((666.7t+32000)^(3/2)/(3/2*666.7)]_0^6#

#=((666.7*6+32000)^(3/2)/(1000))-((666.7*0+32000)^(3/2)/(1000))#

#=36000^(3/2)/1000-32000^(3/2)/1000#

#=1106.2#

So,

#barv=1106.2/6=184.4ms^-1#