An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #18 KJ# to # 64KJ# over #t in [0,12s]#. What is the average speed of the object?

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Answer 1 · 2017-03-15T15:45:09

The average speed of the object is #=199.6ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The initial velocity is #=u_1#

#1/2m u_1^2=18000J#

The final velocity is #=u_2#

#1/2m u_2^2=64000J#

Therefore,

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

and,

#u_2^2=2/2*64000=64000m^2s^-2#

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

The points are #(0,18000)# and #(12,64000)#

The equation of the line is

#v^2-18000=(64000-18000)/12t#

#v^2=3833.3t+18000#

So,

#v=sqrt((3833.3t+18000)#

We need to calculate the average value of #v# over #t in [0,12]#

#(12-0)bar v=int_0^12sqrt((3833.3t+18000)dt#

#12 barv=[((3833.3t+18000)^(3/2)/(3/2*3833.3)]_0^12#

#=((3833.3*12+18000)^(3/2)/(5750))-((3833.3*0+18000)^(3/2)/(5750))#

#=63999.6^(3/2)/5750-18000^(3/2)/5750#

#=2395.8#

So,

#barv=2395.8/12=199.6ms^-1#