An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #12 KJ# to # 72KJ# over #t in [0,6s]#. What is the average speed of the object?

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Answer 1 · 2018-04-03T07:01:37

The average speed is #=163.3ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=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=72000J#

Therefore,

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

and,

#u_2^2=2/3*72000=48000m^2s^-2#

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

The points are #(0,8000)# and #(6,48000)#

The equation of the line is

#v^2-8000=(48000-8000)/6t#

#v^2=6666.7t+8000#

So,

#v=sqrt(6666.7t+8000)#

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

#(6-0)bar v=int_0^6(sqrt(6666.7t+8000))dt#

#6 barv= [(6666.7t+8000)^(3/2)/(3/2*6666.7)] _( 0) ^ (6)#

#=((6666.7*6+8000)^(3/2)/(10000))-((6666.7*0+8000)^(3/2)/(10000))#

#=48000^(3/2)/10000-8000^(3/2)/10000#

#=980.07#

So,

#barv=980.07/6=163.3ms^-1#

The average speed is #=163.3ms^-1#