An object has a mass of #9 kg#. The object's kinetic energy uniformly changes from #54 KJ# to # 0 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

Question

1539 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-12T15:41:55

The average speed is #=73.03ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=9kg#

The initial velocity is #=u_1#

#1/2m u_1^2=54000J#

The final velocity is #=u_2#

#1/2m u_2^2=0J#

Therefore,

#u_1^2=2/9*54000=12000m^2s^-2#

and,

#u_2^2=2/9*0=0m^2s^-2#

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

The points are #(0,12000)# and #(4,0)#

The equation of the line is

#v^2-12000=(0-12000)/4t#

#v^2=-3000t+12000#

So,

#v=sqrt((-3000t+12000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4sqrt((-3000t+12000))dt#

#4 barv=[((-3000t+12000)^(3/2)/(-3/2*3000)]_0^4#

#=((-3000*4+12000)^(3/2)/(-4500))-((-3000*0+12000)^(3/2)/(-4500))#

#=12000^(3/2)/4500-0^(3/2)/4500#

#=292.12#

So,

#barv=292.12/4=73.03ms^-1#