An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #35 KJ# to #18 KJ# over #t in [0, 15 s]#. What is the average speed of the object?

Question

1518 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-13T10:25:00

The average speed is #=162.1ms^-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=35000J#

The final kinetic energy is #1/2m u_2^2=18000J#

Therefore,

#u_1^2=2/2*35000=35000m^2s^-2#

and,

#u_2^2=2/2*18000=18000m^2s^-2#

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

The points are #(0,35000)# and #(15,18000)#

The equation of the line is

#v^2-35000=(18000-35000)/15t#

#v^2=-1133.3t+35000#

So,

#v=sqrt(-1133.3t+35000)#

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

#(15-0)bar v=int_0^15(sqrt(-1133.3t+35000))dt#

#15 barv= (-1133.3t+35000)^(3/2)/(3/2*-1133.3)| _( 0) ^ (15) #

#=((-1133.3*15+35000)^(3/2)/(-1700))-((-1133.3*0+35000)^(3/2)/(-1700))#

#=35000^(3/2)/1700-18000^(3/2)/1700#

#=2431.1#

So,

#barv=2431.1/15=162.1ms^-1#

The average speed is #=162.1ms^-1#