An object has a mass of #3 kg#. The object's kinetic energy uniformly changes from #75 KJ# to #48 KJ# over #t in [0, 15 s]#. What is the average speed of the object?

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Answer 1 · 2018-02-20T10:14:07

The average speed is #=202.1ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=3kg#

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=48000J#

Therefore,

#u_1^2=2/3*75000=50000m^2s^-2#

and,

#u_2^2=2/3*48000=32000m^2s^-2#

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

The points are #(0,50000)# and #(15,32000)#

The equation of the line is

#v^2-50000=(32000-50000)/15t#

#v^2=-1200t+50000#

So,

#v=sqrt(-1200t+50000)#

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

#(15-0)bar v=int_0^15(sqrt(-1200t+50000))dt#

#15 barv= [(-1200t+50000)^(3/2)/(-3/2*1200)] _( 0) ^ (15)#

#=((-1200*15+50000)^(3/2)/(-1800))-((-1200*0+50000)^(3/2)/(-1800))#

#=50000^(3/2)/1800-32000^(3/2)/1800#

#=3031.11#

So,

#barv=3031.11/15=202.1ms^-1#

The average speed is #=202.1ms^-1#