An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #15 KJ# to # 42KJ# over #t in [0,8s]#. What is the average speed of the object?

Question

An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #15 KJ# to # 42KJ# over #t in [0,8s]#.  What is the average speed of the object?

Lily Adams · Accepted Answer

The average speed is #=105.73ms^-1#

The kinetic energy is

#KE=1/2mv^2#

mass is #=5kg#

The initial velocity is #=u_1#

#1/2m u_1^2=15000J#

The final velocity is #=u_2#

#1/2m u_2^2=42000J#

Therefore,

#u_1^2=2/5*15000=6000m^2s^-2#

and,

#u_2^2=2/5*42000=16800m^2s^-2#

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

The points are #(0,6000)# and #(8,16800)#

The equation of the line is

#v^2-6000=(16800-6000)/8t#

#v^2=1350t+6000#

So,

#v=sqrt((1350t+6000)#

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

#(8-0)bar v=int_0^8sqrt((3200t+10666.67))dt#

#8 barv=[((1350t+6000)^(3/2)/(3/2*1350)]_0^8#

#=((1350*8+6000)^(3/2)/(2025))-((1350*0+6000)^(3/2)/(2025))#

#=16800^(3/2)/2025-6000^(3/2)/2025#

#=845.81#

So,

#barv=845.81/8=105.73ms^-1#

Lily Adams · Answer

undefined $$ \text{Average Speed} = \frac{\text{Change in Distance}}{\text{Change in Time}} $$
$$ \text{Average Speed} = \frac{\sqrt{2 \times \text{Change in Kinetic Energy}}}{\text{Mass} \times \text{Change in Time}} $$
$$ \text{Average Speed} = \frac{\sqrt{2 \times (42 \, \text{KJ} - 15 \, \text{KJ})}}{5 \, \text{kg} \times 8 \, \text{s}} $$
$$ \text{Average Speed} \approx \frac{\sqrt{2 \times 27 \, \text{KJ}}}{40 \, \text{kg} \, \text{s}^{-1}} $$
$$ \text{Average Speed} \approx \frac{\sqrt{54} \, \text{KJ}}{40 \, \text{kg} \, \text{s}^{-1}} $$
$$ \text{Average Speed} \approx \frac{7.35 \, \text{KJ}}{40 \, \text{kg} \, \text{s}^{-1}} $$
$$ \text{Average Speed} \approx 0.184 \, \text{KJ} \, \text{kg}^{-1} \, \text{s}^{-1} $$