An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #64 KJ# to #72 KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-09-20T15:08:06

The average speed is #=260.7ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=2kg#

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=64000J#

The final kinetic energy is #1/2m u_2^2=72000J#

Therefore,

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

and,

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

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

The points are #(0,64000)# and #(3,72000)#

The equation of the line is

#v^2-64000=(72000-64000)/3t#

#v^2=2666.7t+64000#

So,

#v=sqrt((2666.7t+64000)#

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

#(3-0)bar v=int_0^3(sqrt(2666.7t+64000))dt#

#3 barv=[((2666.7t+64000)^(3/2)/(3/2*2666.7))]_0^3#

#=((2666.7*3+64000)^(3/2)/(4000))-((2666.7*0+64000)^(3/2)/(4000))#

#=72000^(3/2)/4000-64000^(3/2)/4000#

#=782.2#

So,

#barv=782.2/3=260.7ms^-1#

The average speed is #=260.7ms^-1#