An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #96 KJ# to # 280 KJ# over #t in [0, 12 s]#. What is the average speed of the object?

Question

An object has a mass of #4 kg#. The object's kinetic energy uniformly changes from #96 KJ# to # 280  KJ# over #t in [0,  12  s]#.  What is the average speed of the object?

Elena Albarran · Accepted Answer

The average speed is #=296.65ms^-1#

The initial kinetic energy is #=1/2m u^2#

So,

#1/2m u^2=96000#

#u^2=2*96000/4=48000#

#v=sqrt48000=219.1ms^-1#

The final kinetic energy is #1/2m v^2#

So,

#1/2m v^2=280000#

#v^2=2*280000/4=140000#

#v=sqrt140000=374.2ms^-1#

On a graph velocity v/s time, we have a straight line #DC#

#AD=u#

#BC=v#

#AB=t#

Area of #ABCD=(AD+BC)/2*t#

The average speed #=barv#

#barv*t=(AD+BC)/2*t#

so,

#barv=(u+v)/2#

#=(219.1+374.2)/2#

#=296.65ms^-1#

Adrian Ayala · Answer

undefined To find the average speed of the object, we need to calculate the change in kinetic energy and then use it to find the average speed.

The change in kinetic energy is the final kinetic energy minus the initial kinetic energy: $ \Delta KE = KE_{\text{final}} - KE_{\text{initial}} $.

Given that the initial kinetic energy is 96 kJ and the final kinetic energy is 280 kJ, the change in kinetic energy is:

$ \Delta KE = 280 \, \text{kJ} - 96 \, \text{kJ} = 184 \, \text{kJ} $.

Now, we know that the change in kinetic energy is equal to the work done on the object, which is given by the formula:

$ \Delta KE = \frac{1}{2}mv^2 $, where $ m $ is the mass of the object and $ v $ is its final velocity.

Rearranging the formula to solve for $ v $:

$ v = \sqrt{\frac{2\Delta KE}{m}} $.

Plugging in the values, we get:

$ v = \sqrt{\frac{2 \times 184 \, \text{kJ}}{4 \, \text{kg}}} $.

$ v = \sqrt{\frac{368}{4}} $.

$ v = \sqrt{92} $.

$ v ≈ 9.59 \, \text{m/s} $.

Therefore, the average speed of the object is approximately $ 9.59 \, \text{m/s} $.