An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #15 KJ# to # 64KJ# over #t in [0,12s]#. What is the average speed of the object?

Question

1064 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-19T06:17:04

The average speed is #=107.3ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=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=15000J#

The final kinetic energy is #1/2m u_2^2=64000J#

Therefore,

#u_1^2=2/6*15000=3000m^2s^-2#

and,

#u_2^2=2/6*64000=21333.3m^2s^-2#

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

The points are #(0,3000)# and #(12,21333.3)#

The equation of the line is

#v^2-3000=(21333.3-3000)/12t#

#v^2=1527.8t+3000#

So,

#v=sqrt((1527.8t+3000)#

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

#(12-0)bar v=int_0^12sqrt(1527.8t+3000))dt#

#12 barv=[((1527.8t+3000)^(3/2)/(3/2*1527.8)]_0^12#

#=((1527.8*12+3000)^(3/2)/(2291.7))-((1527.8*0+3000)^(3/2)/(2291.7))#

#=21333.6^(3/2)/2291.7-3000^(3/2)/2291.7#

#=1288#

So,

#barv=1288/12=107.3ms^-1#

The average speed is #=107.3ms^-1#