An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #64 KJ# to # 280 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

Question

1329 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-23T15:37:58

The average speed is #=288.0ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=4kg#

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=280000J#

Therefore,

#u_1^2=2/4*64000=32000m^2s^-2#

and,

#u_2^2=2/4*280000=140000m^2s^-2#

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

The points are #(0,32000)# and #(12,140000)#

The equation of the line is

#v^2-32000=(140000-32000)/12t#

#v^2=9000t+32000#

So,

#v=sqrt((9000t+32000)#

We need to calculate the average value of #v# over #t in [0,12]#

#(12-0)bar v=int_0^12(sqrt(9000t+32000))dt#

#12 barv=[((9000t+32000)^(3/2)/(3/2*9000))]_0^12#

#=((9000*12+32000)^(3/2)/(13500))-((9000*0+32000)^(3/2)/(13500))#

#=140000^(3/2)/13500-32000^(3/2)/13500#

#=3456.2#

So,

#barv=3456.2/12=288.0ms^-1#

The average speed is #=288.0ms^-1#