An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #12 KJ# to # 36KJ# over #t in [0,6s]#. What is the average speed of the object?

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Answer 1 · 2017-08-12T06:27:40

The average speed is #=125.1ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=3kg#

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=12000J#

The final kinetic energy is #1/2m u_2^2=36000J#

Therefore,

#u_1^2=2/3*12000=8000m^2s^-2#

and,

#u_2^2=2/3*36000=24000m^2s^-2#

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

The points are #(0,8000)# and #(6,24000)#

The equation of the line is

#v^2-8000=(24000-8000)/6t#

#v^2=2666.7t+8000#

So,

#v=sqrt((2666.7t+8000)#

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

#(6-0)bar v=int_0^6(sqrt(2666.7t+8000))dt#

#6 barv=[((2666.7t+8000)^(3/2)/(3/2*2666.7))]_0^6#

#=((2666.7*6+8000)^(3/2)/(4000))-((2666.7*0+8000)^(3/2)/(4000))#

#=24000^(3/2)/4000-8000^(3/2)/4000#

#=750.6#

So,

#barv=750.6/6=125.1ms^-1#

The average speed is #=125.1ms^-1#