An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #96 KJ# to # 48 KJ# over #t in [0, 6 s]#. What is the average speed of the object?

Question

1312 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-02-26T06:31:01

The average speed is #=199.7ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=6kg#

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=48000J#

Therefore,

#u_1^2=2/6*96000=32000m^2s^-2#

and,

#u_2^2=2/6*48000=16000m^2s^-2#

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

The points are #(0,48000)# and #(6,32000)#

The equation of the line is

#v^2-48000=(32000-48000)/6t#

#v^2=-2666.7t+48000#

So,

#v=sqrt(-2666.7t+48000)#

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

#(6-0)bar v=int_0^6(sqrt(-2666.7t+48000))dt#

#6 barv= [(-2666.7t+48000)^(3/2)/(-3/2*2666.7)] _( 0) ^ (6)#

#=((-2666.7*6+48000)^(3/2)/(-4000))-((-2666.7*0+48000)^(3/2)/(-4000))#

#=48000^(3/2)/4000-32000^(3/2)/4000#

#=1197.98#

So,

#barv=1197.98/6=199.7ms^-1#

The average speed is #=199.7ms^-1#