An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #48 KJ# to # 36KJ# over #t in [0,6s]#. What is the average speed of the object?

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Answer 1 · 2017-12-30T13:29:17

The average speed is #=167.2ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=3kg#

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=2/3*48000=32000m^2s^-2#

and,

#u_2^2=2/3*36000=24000m^2s^-2#

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

The points are #(0,32000)# and #(6,24000)#

The equation of the line is

#v^2-32000=(24000-32000)/6t#

#v^2=-1333.3t+32000#

So,

#v=sqrt(-1333.3t+32000)#

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

#(6-0)bar v=int_0^6(sqrt(-1333.3t+32000))dt#

#6 barv= (-1333.3t+32000)^(3/2)/(3/2*-1333.3)| _( 0) ^ (6) #

#=((-1333.3*6+32000)^(3/2)/(-2000))-((-1333.3*0+32000)^(3/2)/(-2000))#

#=32000^(3/2)/2000-24000^(3/2)/2000#

#=10003.1#

So,

#barv=1003.1/6=167.2ms^-1#

The average speed is #=167.2ms^-1#