An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #64 KJ# to #38 KJ# over #t in [0, 15 s]#. What is the average speed of the object?

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Answer 1 · 2017-09-12T05:42:14

The average speed is #=225.2ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=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=64000J#

The final kinetic energy is #1/2m u_2^2=38000J#

Therefore,

#u_1^2=2/2*64000=64000m^2s^-2#

and,

#u_2^2=2/2*38000=38000m^2s^-2#

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

The points are #(0,64000)# and #(15,38000)#

The equation of the line is

#v^2-64000=(38000-64000)/15t#

#v^2=-17333t+64000#

So,

#v=sqrt((-1733.3t+64000)#

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

#(15-0)bar v=int_0^15(sqrt(-1733.3t+64000))dt#

#15 barv=[((-1733.3t+64000)^(3/2)/(-3/2*1733.3))]_0^15#

#=((-1733.3*15+64000)^(3/2)/(-2600))-((-1733.3*0+64000)^(3/2)/(-2600))#

#=64000^(3/2)/2600-38000^(3/2)/2600#

#=3378.2#

So,

#barv=3378.2/15=225.2ms^-1#

The average speed is #=225.2ms^-1#