An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #120 KJ# to # 64KJ# over #t in [0,8s]#. What is the average speed of the object?

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Answer 1 · 2018-04-08T07:37:54

The average speed is #=213.6ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=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=120000J#

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

Therefore,

#u_1^2=2/4*120000=60000m^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 #(0,60000)# and #(8,32000)#

The equation of the line is

#v^2-60000=(32000-60000)/8t#

#v^2=-3500t+60000#

So,

#v=sqrt(-3500t+60000)#

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

#(8-0)bar v=int_0^8(sqrt(-3500t+60000))dt#

#8 barv= [(-3500t+60000)^(3/2)/(3/2*-3500)] _( 0) ^ (8)#

#=((-3500*8+60000)^(3/2)/(-5250))-((-3500*0+60000)^(3/2)/(-5250))#

#=60000^(3/2)/5250-32000^(3/2)/5250#

#=1709.1#

So,

#barv=1709.1/8=213.6ms^-1#

The average speed is #=213.6ms^-1#