An object has a mass of #9 kg#. The object's kinetic energy uniformly changes from #54 KJ# to # 45 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

Question

1286 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-02T06:46:59

The average speed is #=104.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=9kg#

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=54000J#

The final kinetic energy is #1/2m u_2^2=45000J#

Therefore,

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

and,

#u_2^2=2/9*45000=10000m^2s^-2#

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

The points are #(0,12000)# and #(4,10000)#

The equation of the line is

#v^2-12000=(10000-12000)/4t#

#v^2=-500t+12000#

So,

#v=sqrt(-500t+12000)#

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

#(4-0)bar v=int_0^4(sqrt(-500t+12000))dt#

#4 barv=[-500t+12000)^(3/2)/(3/2*-500))] _( 0) ^ (4) #

#=((-500*4+12000)^(3/2)/(-750))-((-500*0+12000)^(3/2)/(-750))#

#=12000^(3/2)/750-10000^(3/2)/750#

#=419.4#

So,

#barv=419.4/4=104.8ms^-1#

The average speed is #=104.8ms^-1#