An object has a mass of #12 kg#. The object's kinetic energy uniformly changes from #81 KJ# to # 180 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

Question

1321 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-22T07:37:53

The average speed is #=146.6ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=12kg#

The initial velocity is #=u_1#

#1/2m u_1^2=81000J#

The final velocity is #=u_2#

#1/2m u_2^2=180000J#

Therefore,

#u_1^2=2/12*81000=13500m^2s^-2#

and,

#u_2^2=2/12*180000=30000m^2s^-2#

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

The points are #(0,13500)# and #(4,30000)#

The equation of the line is

#v^2-13500=(30000-13500)/4t#

#v^2=4125t+13500#

So,

#v=sqrt((4125t+13500)#

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

#(4-0)bar v=int_0^4sqrt((4125t+13500))dt#

#4 barv=[((4125t+13500)^(3/2)/(3/2*4125)]_0^4#

#=((4125*4+13500)^(3/2)/(6187.5))-((4125*0+13500)^(3/2)/(6187.5))#

#=30000^(3/2)/6187.5-13500^(3/2)/6187.5#

#=586.3#

So,

#barv=586.3/4=146.6ms^-1#