An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #72 KJ# to # 720 KJ# over #t in [0, 3 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 #72 KJ# to # 720  KJ# over #t in [0,  3  s]#.  What is the average speed of the object?

William Audy · Accepted Answer

The average speed is #=351.4ms^-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=72000J#

The final kinetic energy is #1/2m u_2^2=720000J#

Therefore,

#u_1^2=2/6*72000=24000m^2s^-2#

and,

#u_2^2=2/6*720000=24000m^2s^-2#

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

The points are #(0,24000)# and #(3,240000)#

The equation of the line is

#v^2-24000=(240000-24000)/3t#

#v^2=72000t+24000#

So,

#v=sqrt((72000t+24000)#

We need to calculate the average value of #v# over #t in [0,3]#

#(3-0)bar v=int_0^3sqrt(72000t+24000))dt#

#3 barv=[((72000t+24000)^(3/2)/(3/2*72000)]_0^3#

#=((72000*3+24000)^(3/2)/(108000))-((72000*0+24000)^(3/2)/(108000))#

#=240000^(3/2)/108000-24000^(3/2)/108000#

#=1054.2#

So,

#barv=1054.2/3=351.4ms^-1#

The average speed is #=351.4ms^-1#

Olivia Askew · Answer

undefined The average speed of the object is 8 m/s.

Layla Ahmed · Answer

undefined The average speed of the object can be calculated using the formula:

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

Since the object's kinetic energy uniformly changes, we can assume that its acceleration is constant. Using the equation for kinetic energy:

$$ KE = \frac{1}{2} m v^2 $$

where $ KE $ is the kinetic energy, $ m $ is the mass of the object, and $ v $ is the speed of the object.

Given that the initial kinetic energy is 72 kJ and the final kinetic energy is 720 kJ, we can find the initial and final speeds using these equations:

$$ \text{Initial kinetic energy:} \quad 72 = \frac{1}{2} \times 6 \times v_{\text{initial}}^2 $$

$$ \text{Final kinetic energy:} \quad 720 = \frac{1}{2} \times 6 \times v_{\text{final}}^2 $$

Solving for $ v_{\text{initial}} $ and $ v_{\text{final}} $, we get:

$$ v_{\text{initial}} = \sqrt{\frac{72 \times 2}{6}} $$

$$ v_{\text{final}} = \sqrt{\frac{720 \times 2}{6}} $$

The average speed is then:

$$ \text{Average speed} = \frac{v_{\text{initial}} + v_{\text{final}}}{2} $$