An object has a mass of #12 kg#. The object's kinetic energy uniformly changes from #81 KJ# to # 18 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

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Answer 1 · 2017-03-15T06:58:09

The average speed is #=89.16ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The initial velocity is #=u_1#

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

The final velocity is #=u_2#

#1/2m u_2^2=18000J#

Therefore,

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

and,

#u_2^2=2/12*18000=3000m^2s^-2#

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

The points are #(0,13500)# and #(4,3000)#

The equation of the line is

#v^2-13500=(3000-13500)/4t#

#v^2=-2625t+13500#

So,

#v=sqrt((-2625t+13500)#

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

#(4-0)bar v=int_0^5sqrt(-2625t+13500) dt#

#4 barv=[(-2625t+13500)^(3/2)/(3/2*-2625)]_0^4#

#=((-2625*4+13500)^(3/2)/(-3937.5))-((-2625*0+13500)^(3/2)/(-3937.5))#

#=3000^(3/2)/-3937.5-13500^(3/2)/-3937.5#

#=1/3937.5(13500^(3/2)-3000^(3/2)))#

#=356.63#

So,

#barv=356.63/4=89.16ms^-1#