An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #75 KJ# to #42 KJ# over #t in [0, 15 s]#. What is the average speed of the object?

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Answer 1 · 2017-11-13T07:17:30

The average speed is #=241.1ms^-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=75000J#

The final kinetic energy is #1/2m u_2^2=42000J#

Therefore,

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

and,

#u_2^2=2/2*42000=42000m^2s^-2#

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

The points are #(0,75000)# and #(15,42000)#

The equation of the line is

#v^2-75000=(42000-75000)/15t#

#v^2=-2200t+75000#

So,

#v=sqrt(-2200t+75000)#

We need to calculate the average value of #v# over #t in [0,15]#

#(15-0)bar v=int_0^15(sqrt(-2200t+75000))dt#

#15 barv=[((-2200t+75000)^(3/2)/(-3/2*2200))]_0^15#

#=((-2200*15+75000)^(3/2)/(-3300))-((-2200*0+75000)^(3/2)/(-3300))#

#=75000^(3/2)/3300-42000^(3/2)/3300#

#=3615.8#

So,

#barv=3615.8/15=241.1ms^-1#

The average speed is #=241.1ms^-1#