An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #540 KJ# to # 32 KJ# over #t in [0, 4 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 #540 KJ# to # 32  KJ# over #t in [0,  4  s]#.  What is the average speed of the object?

Benjamin Asiedu · Accepted Answer

The average speed is #=296.3ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=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=540000J#

The final kinetic energy is #1/2m u_2^2=32000J#

Therefore,

#u_1^2=2/6*540000=180000m^2s^-2#

and,

#u_2^2=2/6*32000=10666.7m^2s^-2#

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

The points are #(0,180000)# and #(4,10666.7)#

The equation of the line is

#v^2-180000=(10666.7-180000)/4t#

#v^2=-42333.3t+180000#

So,

#v=sqrt((-42333.3t+180000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4(sqrt(-42333.3t+180000))dt#

#4 barv=[((-42333.3t+180000)^(3/2)/(-3/2*42333.3))]_0^4#

#=((-42333.3*4+180000)^(3/2)/(-63500))-((42333.3*0+180000)^(3/2)/(-63500))#

#=180000^(3/2)/63500-10666.7^(3/2)/63500#

#=1185.3#

So,

#barv=1185.3/4=296.3ms^-1#

The average speed is #=296.3ms^-1#

Leo Abeyta · Answer

undefined To find the average speed of the object, we need to calculate the total distance traveled by the object during the given time interval and then divide it by the total time. We can use the formula for kinetic energy:

$$ KE = \frac{1}{2} mv^2 $$

First, we need to find the initial and final velocities using the given kinetic energies and the mass of the object.

For the initial kinetic energy:

$$ 540 \text{ kJ} = \frac{1}{2} \times 6 \text{ kg} \times v_0^2 $$

$$ v_0^2 = \frac{2 \times 540 \text{ kJ}}{6 \text{ kg}} $$

$$ v_0^2 = 180 \text{ kJ/kg} $$

$$ v_0 = \sqrt{180} \text{ m/s} $$

For the final kinetic energy:

$$ 32 \text{ kJ} = \frac{1}{2} \times 6 \text{ kg} \times v_f^2 $$

$$ v_f^2 = \frac{2 \times 32 \text{ kJ}}{6 \text{ kg}} $$

$$ v_f^2 = 10.67 \text{ kJ/kg} $$

$$ v_f = \sqrt{10.67} \text{ m/s} $$

Now, we can calculate the average speed using the formula:

$$ \text{Average speed} = \frac{\text{Total distance}}{\text{Total time}} $$

Since the object's acceleration is uniform, we can use the average velocity to find the average speed:

$$ \text{Average velocity} = \frac{v_0 + v_f}{2} $$

$$ \text{Average speed} = \frac{2 \times \text{Average velocity}}{\text{Total time}} $$

$$ \text{Average speed} = \frac{2 \times (\frac{\sqrt{180} + \sqrt{10.67}}{2})}{4 \text{ s}} $$

$$ \text{Average speed} = \frac{\sqrt{180} + \sqrt{10.67}}{4} \text{ m/s} $$

$$ \text{Average speed} \approx 6.96 \text{ m/s} $$