An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #16 KJ# to #96 KJ# over #t in [0, 15 s]#. What is the average speed of the object?

Question

1192 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-13T17:06:45

The average speed is #=231.0ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=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=16000J#

The final kinetic energy is #1/2m u_2^2=96000J#

Therefore,

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

and,

#u_2^2=2/2*96000=96000m^2s^-2#

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

The points are #(0,16000)# and #(15,96000)#

The equation of the line is

#v^2-16000=(96000-16000)/15t#

#v^2=5333.3t+16000#

So,

#v=sqrt(5333.3t+16000)#

We need to calculate the average value of #v# over #t in [0,15]#

#(15-0)bar v=int_0^15(sqrt(5333.3t+16000))dt#

#15 barv= (5333.3t+16000)^(3/2)/(3/2*5333.3)| _( 0) ^ (15) #

#=((5333.3*15+16000)^(3/2)/(8000))-((5333.3*0+16000)^(3/2)/(8000))#

#=96000^(3/2)/8000-16000^(3/2)/8000#

#=3465.1#

So,

#barv=3465.1/15=231.0ms^-1#

The average speed is #=231.0ms^-1#