An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #54 KJ# to # 225 KJ# over #t in [0, 8 s]#. What is the average speed of the object?

Question

An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #54 KJ# to # 225  KJ# over #t in [0,  8  s]#.  What is the average speed of the object?

Nathan Ashcroft · Accepted Answer

The average speed is #=212.0ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=6kg#

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=54000J#

The final kinetic energy is #1/2m u_2^2=225000J#

Therefore,

#u_1^2=2/6*54000=18000m^2s^-2#

and,

#u_2^2=2/6*225000=75000m^2s^-2#

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

The points are #(0,18000)# and #(8,75000)#

The equation of the line is

#v^2-18000=(75000-18000)/8t#

#v^2=7125t+18000#

So,

#v=sqrt(7125t+18000)#

We need to calculate the average value of #v# over #t in [0,8]#

#(8-0)bar v=int_0^12(sqrt(7125t+18000))dt#

#8 barv= (7125t+18000)^(3/2)/(3/2*7125)| _( 0) ^ (8) #

#=((7125*8+18000)^(3/2)/(10687.5))-((7125*0+18000)^(3/2)/(10687.5))#

#=75000^(3/2)/10687.5-18000^(3/2)/10687.5#

#=1695.9#

So,

#barv=1695.9/8=212.0ms^-1#

The average speed is #=212.0ms^-1#

Carson Alexander · Answer

undefined To find the average speed of the object, we first need to calculate the change in kinetic energy. The change in kinetic energy (∆KE) is equal to the final kinetic energy minus the initial kinetic energy.

∆KE = 225 KJ - 54 KJ
∆KE = 171 KJ

Next, we need to calculate the average acceleration (a) of the object. The average acceleration is given by the change in kinetic energy divided by the time interval (∆t) over which the change occurs.

a = ∆KE / ∆t
a = 171 KJ / 8 s
a = 21.375 KJ/s

Now, we can use the equation for average speed (v_avg), which is equal to the average acceleration multiplied by the time interval.

v_avg = a * ∆t
v_avg = 21.375 KJ/s * 8 s
v_avg = 171 KJ

So, the average speed of the object is 171 KJ.