An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #225 KJ# to #150 KJ# over #t in [0, 8 s]#. What is the average speed of the object?

2 Answers
Mar 13, 2018

Answer:

The average speed is #=249.6ms^-1#

Explanation:

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=225000J#

The final kinetic energy is #1/2m u_2^2=150000J#

Therefore,

#u_1^2=2/6*225000=75000m^2s^-2#

and,

#u_2^2=2/6*150000=50000m^2s^-2#

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

The points are #(0,75000)# and #(8,50000)#

The equation of the line is

#v^2-75000=(50000-75000)/8t#

#v^2=-3125t+75000#

So,

#v=sqrt(-3125t+75000)#

We need to calculate the average value of #v# over #t in [0,8]#

#(8-0)bar v=int_0^8(sqrt(-3125t+75000))dt#

#8 barv= [(-3125t+75000)^(3/2)/(-3/2*3125)] _( 0) ^ (8)#

#=((-3125*8+75000)^(3/2)/(-4687.5))-((-3125*0+75000)^(3/2)/(-4687.5))#

#=75000^(3/2)/4687.5-50000^(3/2)/4687.5#

#=1996.6#

So,

#barv=1996.6/8=249.6ms^-1#

The average speed is #=249.6ms^-1#

Mar 14, 2018

Answer:

#248.73 m/s#

Explanation:

#KE=1/2mv^2#

#v=sqrt((2KE)/m)#

So, v initial #=(sqrt((2*225,000J)/(6 kg))=273.86 m/s#

and v final#=(sqrt((2*150,000J)/(6 kg))=223.607 m/s#

#v avg=((v f+v i)/2)=((273.86 + 223.607 m/s)/2)=248.73 m/s#