An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #480 KJ# to #108 KJ# over #t in [0, 8 s]#. What is the average speed of the object?

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Answer 1 · 2017-06-07T08:39:16

The average speed is #=307.4ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=6kg#

The initial velocity is #=u_1#

The initial kinetic energy is #1/2m u_1^2=480000J#

The final velocity is #=u_2#

The final kinetic energy is #1/2m u_2^2=108000J#

Therefore,

#u_1^2=2/6*480000=160000m^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 #(0,160000)# and #(8,36000)#

The equation of the line is

#v^2-160000=(36000-160000)/8t#

#v^2=-15500t+160000#

So,

#v=sqrt((-15500t+160000)#

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

#(8-0)bar v=int_0^8sqrt((-15500t+160000))dt#

#8 barv=[((-15500t+160000)^(3/2)/(-3/2*15500)]_0^8#

#=((-15500*8+160000)^(3/2)/(-23250))-((-15500*0+160000)^(3/2)/(-23250))#

#=160000^(3/2)/23250-36000^(3/2)/23250#

#=2458.9#

So,

#barv=2458.9/8=307.4ms^-1#

The average speed is #=307.4ms^-1#