An object has a mass of #1 kg#. The object's kinetic energy uniformly changes from #48 KJ# to #36 KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2018-01-30T16:27:21

The average speed is #=204.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=1kg#

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=48000J#

The final kinetic energy is #1/2m u_2^2=36000J#

Therefore,

#u_1^2=2/1*48000=96000m^2s^-2#

and,

#u_2^2=2/1*36000=72000m^2s^-2#

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

The points are #(0,48000)# and #(3,36000)#

The equation of the line is

#v^2-48000=(36000-48000)/3t#

#v^2=-4000t+48000#

So,

#v=sqrt(-4000t+48000)#

We need to calculate the average value of #v# over #t in [0,3]#

#(3-0)bar v=int_0^3(sqrt(-4000t+48000))dt#

#3 barv= [(-4000t+48000)^(3/2)/(3/2*-4000)] _( 0) ^ (3)#

#=((-4000*3+48000)^(3/2)/(-6000))-((-4000*0+48000)^(3/2)/(-6000))#

#=48000^(3/2)/6000-36000^(3/2)/6000#

#=614.3#

So,

#barv=614.3/3=204.8ms^-1#

The average speed is #=204.8ms^-1#