An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #120 KJ# to # 64KJ# over #t in [0,6s]#. What is the average speed of the object?

Question

1198 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-09-03T08:51:12

The average speed is #=213.6ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=4kg#

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=120000J#

The final kinetic energy is #1/2m u_2^2=64000J#

Therefore,

#u_1^2=2/4*120000=60000m^2s^-2#

and,

#u_2^2=2/4*64000=32000m^2s^-2#

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

The points are #(0,60000)# and #(6,32000)#

The equation of the line is

#v^2-60000=(32000-60000)/6t#

#v^2=-4666.6t+60000#

So,

#v=sqrt((-4666.7t+60000)#

We need to calculate the average value of #v# over #t in [0,6]#

#(6-0)bar v=int_0^6(sqrt(-4666.7t+60000))dt#

#6 barv=[((-4666.7t+60000)^(3/2)/(-3/2*3500))]_0^6#

#=((-4666.7*6+60000)^(3/2)/(-7000))-((-4666.7*0+60000)^(3/2)/(-7000))#

#=60000^(3/2)/7000-32000^(3/2)/7000#

#=1281.8#

So,

#barv=1281.8/6=213.6ms^-1#

The average speed is #=213.6ms^-1#