An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #160 KJ# to # 36KJ# over #t in [0, 6 s]#. What is the average speed of the object?

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Answer 1 · 2018-05-21T11:51:05

The average speed is #=307.4ms^-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=160000J#

The final kinetic energy is #1/2m u_2^2=36000J#

Therefore,

#u_1^2=2/2*160000=160000m^2s^-2#

and,

#u_2^2=2/2*36000=36000m^2s^-2#

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

The points are #(0,160000)# and #(6, 36000)#

The equation of the line is

#v^2-160000=(36000-160000)/6t#

#v^2=-20666.7t+160000#

So,

#v=sqrt(-20666.7t+160000)#

We need to calculate the average value of #v# over #t in [0,6]#

#(6-0)bar v=int_0^6(sqrt(-20666.7t+160000))dt#

#6 barv= [(-20666.7t+160000)^(3/2)/(3/2*-20666.7)] _( 0) ^ (6)#

#=((-20666.7*6+160000)^(3/2)/(-31000))-((-20666.7*0+160000)^(3/2)/(-31000))#

#=160000^(3/2)/31000-36000^(3/2)/31000#

#=1844.2#

So,

#barv=1844.2/6=307.4ms^-1#

The average speed is #=307.4ms^-1#