An object has a mass of #5 kg#. The object's kinetic energy uniformly changes from #15 KJ# to # 64KJ# over #t in [0,12s]#. 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 # 64KJ# over #t in [0,12s]#.  What is the average speed of the object?

Elena Albarran · Accepted Answer

The average speed is #=123.5ms^-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=64000J#

Therefore,

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

and,

#u_2^2=2/5*64000=25600m^2s^-2#

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

The points are #(0,6000)# and #(12,25600)#

The equation of the line is

#v^2-6000=(25600-6000)/12t#

#v^2=1633.3t+6000#

So,

#v=sqrt((1633.3t+6000)#

We need to calculate the average value of #v# over #t in [0,12]#

#(12-0)bar v=int_0^12sqrt((1633.3t+6000))dt#

#12 barv=[((1633.3t+6000)^(3/2)/(3/2*1633.3)]_0^12#

#=((1633.3*12+6000)^(3/2)/(2450))-((1633.3*0+6000)^(3/2)/(2450))#

#=25600^(3/2)/2450-6000^(3/2)/2450#

#=1482.1#

So,

#barv=1482.1/12=123.5ms^-1#

Elena Albarran · Answer

undefined The average speed of the object is calculated by dividing the total distance traveled by the object by the total time taken. In this case, since the kinetic energy uniformly changes from 15 kJ to 64 kJ over 12 seconds, we need to find the average kinetic energy of the object during this time period. Then, we can use the formula for kinetic energy to find the average speed of the object. Given that kinetic energy $ E_k = \frac{1}{2} mv^2 $, where $ m $ is the mass of the object and $ v $ is its velocity, we can rearrange the formula to find the velocity $ v $ as $ v = \sqrt{\frac{2E_k}{m}} $. We'll use the average kinetic energy $ \bar{E_k} $ during the time interval.

Given:
Initial kinetic energy $ E_{k1} = 15 $ kJ
Final kinetic energy $ E_{k2} = 64 $ kJ
Mass of the object $ m = 5 $ kg
Time interval $ t = 12 $ s

Average kinetic energy $ \bar{E_k} = \frac{E_{k1} + E_{k2}}{2} $

Average velocity $ \bar{v} = \sqrt{\frac{2\bar{E_k}}{m}} $

$$ \bar{E_k} = \frac{15 + 64}{2} = 39.5 \text{ kJ} $$

$$ \bar{v} = \sqrt{\frac{2 \times 39.5 \text{ kJ}}{5 \text{ kg}}} $$

$$ \bar{v} = \sqrt{\frac{79 \text{ kJ}}{5 \text{ kg}}} $$

$$ \bar{v} = \sqrt{15.8} \text{ m/s} \approx 3.98 \text{ m/s} $$

Therefore, the average speed of the object is approximately 3.98 m/s.