An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #64 KJ# to #28 KJ# over #t in [0, 15 s]#. What is the average speed of the object?

Question

1344 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-06T15:12:17

The average speed is #=319.6ms^-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=64000J#

The final kinetic energy is #1/2m u_2^2=28000J#

Therefore,

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

and,

#u_2^2=2/2*28000=28000m^2s^-2#

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

The points are #(0,64000)# and #(15,28000)#

The equation of the line is

#v^2-64000=(28000-64000)/15t#

#v^2=-2400t+64000#

So,

#v=sqrt(-2400t+64000)#

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

#(15-0)bar v=int_0^15(sqrt(-2400t+64000))dt#

#15 barv= (-2400t+64000)^(3/2)/(3/2*-2400)| _( 0) ^ (15) #

#=((-2400*15+64000)^(3/2)/(-2400))-((-2400*0+64000)^(3/2)/(-2400))#

#=64000^(3/2)/2400-28000^(3/2)/2400#

#=4794#

So,

#barv=4794/15=319.6ms^-1#

The average speed is #=319.6ms^-1#