An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #254 KJ# to # 24 KJ# over #t in [0, 5 s]#. What is the average speed of the object?

Question

1255 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-04-13T06:54:57

The average speed is #=254.75ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=4kg#

The initial velocity is #=u_1#

#1/2m u_1^2=254000J#

The final velocity is #=u_2#

#1/2m u_2^2=24000J#

Therefore,

#u_1^2=2/4*254000=127000m^2s^-2#

and,

#u_2^2=2/4*24000=12000m^2s^-2#

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

The points are #(0,127000)# and #(5,12000)#

The equation of the line is

#v^2-127000=(12000-127000)/5t#

#v^2=-23000t+127000#

So,

#v=sqrt((-23000t+127000)#

We need to calculate the average value of #v# over #t in [0,5]#

#(5-0)bar v=int_0^5sqrt((-23000t+127000))dt#

#5 barv=[((-23000t+127000)^(3/2)/(-3/2*23000)]_0^5#

#=((-23000*5+127000)^(3/2)/(-34500))-((-23000*0+127000)^(3/2)/(-34500))#

#=127000^(3/2)/34500-12000^(3/2)/34500#

#=1273.75#

So,

#barv=1273.75/5=254.75ms^-1#