An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #72 KJ# to # 720 KJ# over #t in [0, 3 s]#. What is the average speed of the object?

Question

1142 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-20T18:22:48

The average speed is #=351.4ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=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=72000J#

The final kinetic energy is #1/2m u_2^2=720000J#

Therefore,

#u_1^2=2/6*72000=24000m^2s^-2#

and,

#u_2^2=2/6*720000=24000m^2s^-2#

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

The points are #(0,24000)# and #(3,240000)#

The equation of the line is

#v^2-24000=(240000-24000)/3t#

#v^2=72000t+24000#

So,

#v=sqrt((72000t+24000)#

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

#(3-0)bar v=int_0^3sqrt(72000t+24000))dt#

#3 barv=[((72000t+24000)^(3/2)/(3/2*72000)]_0^3#

#=((72000*3+24000)^(3/2)/(108000))-((72000*0+24000)^(3/2)/(108000))#

#=240000^(3/2)/108000-24000^(3/2)/108000#

#=1054.2#

So,

#barv=1054.2/3=351.4ms^-1#

The average speed is #=351.4ms^-1#