An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #630 KJ# to # 360 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

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Answer 1 · 2018-01-18T17:48:37

The average speed is #=404.9ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=6kg#

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=630000J#

The final kinetic energy is #1/2m u_2^2=360000J#

Therefore,

#u_1^2=2/6*630000=210000m^2s^-2#

and,

#u_2^2=2/6*360000=120000m^2s^-2#

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

The points are #(0,210000)# and #(4,120000)#

The equation of the line is

#v^2-210000=(120000-210000)/4t#

#v^2=-22500t+210000#

So,

#v=sqrt(-22500t+210000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4(sqrt(-22500t+210000))dt#

#4 barv= (-22500t+210000)^(3/2)/(3/2*-22500)| _( 0) ^ (4) #

#=((-22500*4+210000)^(3/2)/(-33750))-((-22500*0+210000)^(3/2)/(-33750))#

#=210000^(3/2)/33750-120000^(3/2)/33750#

#=1619.7#

So,

#barv=1619.7/4=404.9ms^-1#

The average speed is #=404.9ms^-1#