An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #540 KJ# to # 32 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

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Answer 1 · 2017-07-24T17:15:12

The average speed is #=296.3ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=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=540000J#

The final kinetic energy is #1/2m u_2^2=32000J#

Therefore,

#u_1^2=2/6*540000=180000m^2s^-2#

and,

#u_2^2=2/6*32000=10666.7m^2s^-2#

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

The points are #(0,180000)# and #(4,10666.7)#

The equation of the line is

#v^2-180000=(10666.7-180000)/4t#

#v^2=-42333.3t+180000#

So,

#v=sqrt((-42333.3t+180000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4(sqrt(-42333.3t+180000))dt#

#4 barv=[((-42333.3t+180000)^(3/2)/(-3/2*42333.3))]_0^4#

#=((-42333.3*4+180000)^(3/2)/(-63500))-((42333.3*0+180000)^(3/2)/(-63500))#

#=180000^(3/2)/63500-10666.7^(3/2)/63500#

#=1185.3#

So,

#barv=1185.3/4=296.3ms^-1#

The average speed is #=296.3ms^-1#