An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #128 KJ# to # 720 KJ# over #t in [0, 3 s]#. What is the average speed of the object?

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Answer 1 · 2017-05-21T06:36:09

The average speed is #=367.4ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=6kg#

The initial velocity is #=u_1#

#1/2m u_1^2=128000J#

The final velocity is #=u_2#

#1/2m u_2^2=720000J#

Therefore,

#u_1^2=2/6*128000=42666.7m^2s^-2#

and,

#u_2^2=2/6*720000=240000m^2s^-2#

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

The points are #(0,42666.7)# and #(3,240000)#

The equation of the line is

#v^2-42666.7=(240000-42666.7)/3t#

#v^2=65777.8t+42666.7#

So,

#v=sqrt((65777.8t+42666.7)#

We need to calculate the average value of #v# over #t in [0,3]#

#(3-0)bar v=int_0^8sqrt((65777.8t+42666.7))dt#

#3 barv=[((65777.8t+42666.7)^(3/2)/(3/2*65666.7)]_0^3#

#=((65777.8*3+42666.7)^(3/2)/(98666.7))-((65777.8*0+42666.7)^(3/2)/(98666.7))#

#=240000^(3/2)/98666.7-42666.7^(3/2)/98666.7#

#=1102.3#

So,

#barv=1103.2/3=367.4ms^-1#