An object has a mass of #8 kg#. The object's kinetic energy uniformly changes from #240 KJ# to # 64KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-06-23T09:45:07

The average speed is #=192ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=8kg#

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=240000J#

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

Therefore,

#u_1^2=2/8*240000=60000m^2s^-2#

and,

#u_2^2=2/8*64000=16000m^2s^-2#

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

The points are #(0,60000)# and #(3,16000)#

The equation of the line is

#v^2-60000=(16000-60000)/3t#

#v^2=-14666.7t+60000#

So,

#v=sqrt((-14666.7t+60000)#

We need to calculate the average value of #v# over #t in [0,3]#

#(3-0)bar v=int_0^3sqrt(-14666.7t+60000))dt#

#3 barv=[((-14666.7t+60000)^(3/2)/(-3/2*14666.7)]_0^3#

#=((-14666.7*3+60000)^(3/2)/(22000))-((14666.7*0+60000)^(3/2)/(22000))#

#=60000^(3/2)/22000-16000^(3/2)/22000#

#=576#

So,

#barv=576/3=192ms^-1#

The average speed is #=192ms^-1#