An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #55 KJ# to # 24 KJ# over #t in [0,4s]#. What is the average speed of the object?

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Answer 1 · 2017-04-20T08:45:57

The average speed is #=124.9ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=5kg#

The initial velocity is #=u_1#

#1/2m u_1^2=55000J#

The final velocity is #=u_2#

#1/2m u_2^2=24000J#

Therefore,

#u_1^2=2/5*55000=22000m^2s^-2#

and,

#u_2^2=2/5*24000=9600m^2s^-2#

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

The points are #(0,22000)# and #(4,9600)#

The equation of the line is

#v^2-22000=(9600-22000)/4t#

#v^2=-3100t+22000#

So,

#v=sqrt((-3100t+22000)#

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

#(4-0)bar v=int_0^4sqrt((-3100t+22000))dt#

#4 barv=[((-3100t+22000)^(3/2)/(-3/2*3100)]_0^4#

#=((-3100*4+22000)^(3/2)/(-4650))-((-3100*0+22000)^(3/2)/(-4650))#

#=22000^(3/2)/4650-9600^(3/2)/4650#

#=499.5#

So,

#barv=499.5/4=124.9ms^-1#