An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #128 KJ# to # 48 KJ# over #t in [0, 5 s]#. What is the average speed of the object?

Question

1401 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-12T09:11:16

The average speed is #=207.9ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=4kg#

The initial velocity is #=u_1#

The initial kinetic energy is #1/2m u_1^2=128000J#

The final velocity is #=u_2#

The final kinetic energy is #1/2m u_2^2=48000J#

Therefore,

#u_1^2=2/4*128000=64000m^2s^-2#

and,

#u_2^2=2/4*48000=24000m^2s^-2#

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

The points are #(0,64000)# and #(5,24000)#

The equation of the line is

#v^2-64000=(24000-64000)/5t#

#v^2=-8000t+64000#

So,

#v=sqrt((-8000t+64000)#

We need to calculate the average value of #v# over #t in [0,5]#

#(5-0)bar v=int_0^5sqrt((-8000t+64000))dt#

#5 barv=[((-8000t+64000)^(3/2)/(3/2*8000)]_0^5#

#=((-8000*5+64000)^(3/2)/(12000))-((-8000*0+64000)^(3/2)/(12000))#

#=64000^(3/2)/12000-24000^(3/2)/12000#

#=1039.4#

So,

#barv=1039.4/5=207.9ms^-1#

The average speed is #=207.9ms^-1#