An object has a mass of #12 kg#. The object's kinetic energy uniformly changes from #96 KJ# to # 160 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

Question

1250 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-06-19T18:42:38

The average speed is #=145.7ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=12kg#

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=96000J#

The final kinetic energy is #1/2m u_2^2=160000J#

Therefore,

#u_1^2=2/12*96000=16000m^2s^-2#

and,

#u_2^2=2/12*160000=26666.7m^2s^-2#

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

The points are #(0,16000)# and #(4,26666.7)#

The equation of the line is

#v^2-16000=(26666.7-16000)/4t#

#v^2=2666.7t+16000#

So,

#v=sqrt((2666.7t+16000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4sqrt((2666.7t+16000))dt#

#4 barv=[((2666.7t+16000)^(3/2)/(3/2*2666.7)]_0^4#

#=((2666.7*4+16000)^(3/2)/(4000))-((2666.7*0+16000)^(3/2)/(4000))#

#=26666.7^(3/2)/4000-16000^(3/2)/4000#

#=582.7#

So,

#barv=582.7/4=145.7ms^-1#

The average speed is #=145.7ms^-1#