An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #16 KJ# to # 66KJ# over #t in [0, 6 s]#. What is the average speed of the object?

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Answer 1 · 2018-06-15T17:48:04

The average speed is #=140.8ms^-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=16000J#

The final kinetic energy is #1/2m u_2^2=66000J#

Therefore,

#u_1^2=2/4*16000=8000m^2s^-2#

and,

#u_2^2=2/4*66000=33000m^2s^-2#

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

The points are #(0,8000)# and #(6, 33000)#

The equation of the line is

#v^2-8000=(33000-8000)/6t#

#v^2=4166.7t+8000#

So,

#v=sqrt(4166.7t+8000)#

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

#(6-0)bar v=int_0^6(sqrt(4166.7t+8000))dt#

#6 barv= [(4166.7t+8000)^(3/2)/(3/2*4166.7)] _( 0) ^ (6)#

#=((4166.7*6+8000)^(3/2)/(6250))-((4166.7*0+8000)^(3/2)/(6250))#

#=33000^(3/2)/6250-8000^(3/2)/6250#

#=844.67#

So,

#barv=844.67/6=140.8ms^-1#

The average speed is #=140.8ms^-1#