An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #150 KJ# to # 64KJ# over #t in [0,8s]#. What is the average speed of the object?

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Answer 1 · 2017-07-12T05:43:50

The average speed is #=229.7ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=4kg#

The initial velocity is #=u_1ms^-1#

The final velocity is #=u_2 ms^-1#

The initial kinetic energy is #1/2m u_1^2=150000J#

The final kinetic energy is #1/2m u_2^2=64000J#

Therefore,

#u_1^2=2/4*150000=75000m^2s^-2#

and,

#u_2^2=2/4*64000=32000m^2s^-2#

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

The points are #(0,75000)# and #(8,32000)#

The equation of the line is

#v^2-75000=(32000-75000)/8t#

#v^2=-5375t+75000#

So,

#v=sqrt((-5375t+75000)#

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

#(8-0)bar v=int_0^8sqrt(-5375t+75000))dt#

#8 barv=[((-5375t+75000)^(3/2)/(-3/2*5375)]_0^8#

#=((-5375*8+75000)^(3/2)/(-8062.5))-((-5375*0+75000)^(3/2)/(-8062.5))#

#=75000^(3/2)/8062.5-32000^(3/2)/8062.5#

#=1837.6#

So,

#barv=1837.6/8=229.7ms^-1#

The average speed is #=229.7ms^-1#