An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #540 KJ# to # 36 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

Question

971 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-10T14:54:13

The average speed is #=297.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=6kg#

The initial velocity is #=u_1#

#1/2m u_1^2=540000J#

The final velocity is #=u_2#

#1/2m u_2^2=36000J#

Therefore,

#u_1^2=2/6*540000=180000m^2s^-2#

and,

#u_2^2=2/6*36000=12000m^2s^-2#

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

The points are #(0,180000)# and #(4,12000)#

The equation of the line is

#v^2-180000=(12000-180000)/4t#

#v^2=-42000t+180000#

So,

#v=sqrt((-42000t+180000)#

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

#(4-0)bar v=int_0^4sqrt((-42000t+180000))dt#

#4 barv=[((-42000t+180000)^(3/2)/(-3/2*42000)]_0^4#

#=((-42000*4+180000)^(3/2)/(-63000))-((-42000*0+180000)^(3/2)/(-63000))#

#=180000^(3/2)/63000-12000^(3/2)/63000#

#=1191.3#

So,

#barv=1191.3/4=297.8ms^-1#