An object has a mass of #12 kg#. The object's kinetic energy uniformly changes from #254 KJ# to # 24 KJ# over #t in [0, 5 s]#. What is the average speed of the object?

Question

An object has a mass of #12 kg#. The object's kinetic energy uniformly changes from #254 KJ# to # 24  KJ# over #t in [0,  5  s]#.  What is the average speed of the object?

Lily Adams · Accepted Answer

The average speed is #=147.1ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #m=12kg#

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=254000J#

The final kinetic energy is #1/2m u_2^2=24000J#

Therefore,

#u_1^2=2/12*254000=42333.3m^2s^-2#

and,

#u_2^2=2/12*24000=4000m^2s^-2#

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

The points are #(0,42333.3)# and #(5,4000)#

The equation of the line is

#v^2-42333.3=(4000-42333.3)/5t#

#v^2=-7666.7t+42333.3#

So,

#v=sqrt(-7666.7t+42333.3)#

We need to calculate the average value of #v# over #t in [0,5]#

#(5-0)bar v=int_0^5(sqrt(-7666.7t+42333.3))dt#

#5 barv= (-7666.7t+42333.3)^(3/2)/(3/2*-7666.7)| _( 0) ^ (5) #

#=((-7666.7*5+42333.3)^(3/2)/(-11500))-((-7666.7*0+42333.3)^(3/2)/(-11500))#

#=42333.3^(3/2)/11500-4000^(3/2)/11500#

#=735.4#

So,

#barv=735.4/5=147.1ms^-1#

The average speed is #=147.1ms^-1#

Xavier Adame · Answer

undefined To find the average speed of the object, first, we need to calculate the initial and final velocities using the given kinetic energy values and the formula:

$$ \text{Kinetic energy} = \frac{1}{2}mv^2 $$

Then, we can use the average speed formula:

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

Since the acceleration is not given explicitly, we'll use the equations of motion to find the final velocity. We'll assume constant acceleration, which is not the case in all situations, but it's a common assumption for problems like this. The equation we'll use is:

$$ v_f = v_i + at $$

Given that the initial velocity is 0 (since the object starts from rest), we can rearrange this equation to:

$$ v_f = at $$

Then, we'll use the formula for kinetic energy to find the final velocity:

$$ K.E = \frac{1}{2}mv_f^2 $$

$$ v_f = \sqrt{\frac{2K.E}{m}} $$

Now, we can find the acceleration:

$$ a = \frac{v_f - v_i}{t} $$

Finally, we can use the average speed formula:

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

Given that the object starts from rest, the total distance traveled is the average distance traveled, which can be calculated using the average velocity:

$$ \text{Total distance} = \text{Average velocity} \times \text{Total time} $$

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

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

Substituting the known values:

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

$$ \text{Average speed} = \frac{v_f}{2} \times 5 $$

$$ \text{Average speed} = \frac{\sqrt{\frac{2K.E}{m}}}{2} \times 5 $$

$$ \text{Average speed} = \frac{\sqrt{\frac{2 \times 24 \times 10^3}{12}}}{2} \times 5 $$

$$ \text{Average speed} = \frac{\sqrt{4000}}{2} \times 5 $$

$$ \text{Average speed} = \frac{63.25}{2} \times 5 $$

$$ \text{Average speed} ≈ 157.5 \, \text{m/s} $$