An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #84 KJ# to # 124KJ# over #t in [0, 6 s]#. What is the average speed of the object?

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Answer 1 · 2017-03-27T12:20:07

The average speed is #=227.7ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=84000J#

The final velocity is #=u_2#

#1/2m u_2^2=124000J#

Therefore,

#u_1^2=2/4*84000=42000m^2s^-2#

and,

#u_2^2=2/4*124000=62000m^2s^-2#

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

The points are #(0,42000)# and #(6,62000)#

The equation of the line is

#v^2-42000=(62000-42000)/6t#

#v^2=3333.3t+42000#

So,

#v=sqrt((3333.3t+42000)#

We need to calculate the average value of #v# over #t in [0,6]#

#(6-0)bar v=int_0^6sqrt((3333.3t+42000)dt#

#6 barv=[((3333.3t+42000)^(3/2)/(3/2*3333.3)]_0^3#

#=((3333.3*6+42000)^(3/2)/(5000))-((3333.3*0+42000)^(3/2)/(5000))#

#=62000^(3/2)/5000-42000^(3/2)/5000#

#=1366.1#

So,

#barv=1366.1/6=227.7ms^-1#