An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #75 KJ# to #135 KJ# over #t in [0, 9 s]#. What is the average speed of the object?

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Answer 1 · 2017-10-29T13:31:17

The average speed is #=204.2ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=5kg#

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=75000J#

The final kinetic energy is #1/2m u_2^2=135000J#

Therefore,

#u_1^2=2/5*75000=30000m^2s^-2#

and,

#u_2^2=2/5*135000=54000m^2s^-2#

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

The points are #(0,30000)# and #(9,54000)#

The equation of the line is

#v^2-30000=(54000-30000)/9t#

#v^2=2666.7t+30000#

So,

#v=sqrt((2666.7t+30000)#

We need to calculate the average value of #v# over #t in [0,9]#

#(9-0)bar v=int_0^9(sqrt(2666.7t+30000))dt#

#9 barv=[((2666.7t+30000)^(3/2)/(3/2*2666.7))]_0^9#

#=((2666.7*9+30000)^(3/2)/(4000))-((2666.7*0+30000)^(3/2)/(4000))#

#=54000^(3/2)/4000-30000^(3/2)/4000#

#=1838.1#

So,

#barv=1838.1/9=204.2ms^-1#

The average speed is #=204.2ms^-1#