An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #42 KJ# to # 15 KJ# over #t in [0, 8 s]#. What is the average speed of the object?

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Answer 1 · 2017-04-06T06:01:54

The average speed is #=96.5ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=6kg#

The initial velocity is #=u_1#

#1/2m u_1^2=42000J#

The final velocity is #=u_2#

#1/2m u_2^2=15000J#

Therefore,

#u_1^2=2/6*42000=14000m^2s^-2#

and,

#u_2^2=2/6*15000=5000m^2s^-2#

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

The points are #(0,14000)# and #(8,5000)#

The equation of the line is

#v^2-14000=(5000-14000)/8t#

#v^2=-1125t+14000#

So,

#v=sqrt((-1125t+14000)#

We need to calculate the average value of #v# over #t in [0,8]#

#(8-0)bar v=int_0^8sqrt((-1125t+14000))dt#

#8 barv=[((-1125t+14000)^(3/2)/(-3/2*1125)]_0^8#

#=((-1125*8+14000)^(3/2)/(-1687.5))-((-1125*0+14000)^(3/2)/(-1687.5))#

#=-5000^(3/2)/1685.5+14000^(3/2)/1687.5#

#=772.1#

So,

#barv=772.1/8=96.5ms^-1#