An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #120 KJ# to # 64KJ# over #t in [0,5s]#. What is the average speed of the object?

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Answer 1 · 2018-04-04T11:37:58

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 #(5,32000)#

The equation of the line is

#v^2-60000=(32000-60000)/5t#

#v^2=-5600t+60000#

So,

#v=sqrt(-5600t+60000)#

We need to calculate the average value of #v# over #t in [0,5]#

#(5-0)bar v=int_0^5(sqrt(-5600t+60000))dt#

#5 barv= [(-5600t+60000)^(3/2)/(3/2*-5600)] _( 0) ^ (5)#

#=((-5600*5+60000)^(3/2)/(-8400))-((-5600*0+60000)^(3/2)/(-8400))#

#=60000^(3/2)/8400-32000^(3/2)/8400#

#=1068.2#

So,

#barv=1068.2/5=213.6ms^-1#

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