An object has a mass of #12 kg#. The object's kinetic energy uniformly changes from #64 KJ# to # 140 KJ# over #t in [0, 5 s]#. What is the average speed of the object?

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Answer 1 · 2017-03-13T10:39:55

The average speed is #=129.6ms^-1#

The kinetis energy is

#KE=1/2mv^2#

The initial velocity is #=u_1#

#1/2m u_1^2=64000J#

The final velocity is #=u_2#

#1/2m u_2^2=140000J#

Therefore,

#u_1^2=2/12*64000=10666.7m^2s^-2#

and,

#u_2^2=2/12*140000=23333.3m^2s^-2#

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

The points are #(0,1066.7)# and #(5,23333.3)#

The equation of the line is

#v^2-10666.7=(23333.3-10666.7)/5t#

#v^2=(12666.6)/5t+10666.7#

So,

#v=sqrt((12666.6)/5t+10666.7)#

We need to calculate the average value of #v# over #t in [0,5]#

#(5-0)bar v=int_0^5sqrt((12666.6)/5t+10666.7)dt#

#5 barv=[((12666.6)/5t+10666.7)^(3/2)/(3/2*12666.6/5)]_0^5#

#=(((12666.6)/5*5+10666.7)^(3/2)/(3800))-(((12666.6)/5*0+10666.7)^(3/2)/(3800))#

#=23333.3^(3/2)/3800-10666.7^(3/2)/3800#

#=648#

So,

#barv=648/5=129.6ms^-1#