An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #120 KJ# to # 720 KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-12-12T10:57:36

The average speed is #=365.3ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=6kg#

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=720000J#

Therefore,

#u_1^2=2/6*120000=40000m^2s^-2#

and,

#u_2^2=2/6*720000=240000m^2s^-2#

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

The points are #(0,40000)# and #(3,240000)#

The equation of the line is

#v^2-40000=(240000-40000)/3t#

#v^2=66666.7t+40000#

So,

#v=sqrt(66666.7t+40000)#

We need to calculate the average value of #v# over #t in [0,3]#

#(3-0)bar v=int_0^3(sqrt(66666.7t+40000))dt#

#3 barv= [ 66666.7t+40000)^(3/2)/(3/2*66666.7)| _( 0) ^ (3) #

#=((66666.7*3+40000)^(3/2)/(100000))-((66666.7*0+40000)^(3/2)/(100000))#

#=240000^(3/2)/100000-40000^(3/2)/100000#

#=1095.76#

So,

#barv=1095.76/3=365.3ms^-1#

The average speed is #=365.3ms^-1#