An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #84 KJ# to # 96KJ# over #t in [0, 6 s]#. What is the average speed of the object?

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Answer 1 · 2017-10-29T05:50:52

The average speed is #=212.1ms^-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=84000J#

The final kinetic energy is #1/2m u_2^2=96000J#

Therefore,

#u_1^2=2/4*84000=42000m^2s^-2#

and,

#u_2^2=2/4*96000=48000m^2s^-2#

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

The points are #(0,42000)# and #(6,48000)#

The equation of the line is

#v^2-42000=(48000-42000)/6t#

#v^2=1000t+42000#

So,

#v=sqrt((1000t+42000)#

We need to calculate the average value of #v# over #t in [0,6]#

#(16-0)bar v=int_0^6(sqrt(1000t+42000))dt#

#6 barv=[((1000t+42000)^(3/2)/(3/2*1000))]_0^6#

#=((1000*6+42000)^(3/2)/(1500))-((1000*0+42000)^(3/2)/(1500))#

#=48000^(3/2)/1500-42000^(3/2)/1500#

#=1272.6#

So,

#barv=1272.6/6=212.1ms^-1#

The average speed is #=212.1ms^-1#