An object has a mass of #6 kg#. The object's kinetic energy uniformly changes from #84 KJ# to # 12 KJ# over #t in [0, 6 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 #84 KJ# to # 12  KJ# over #t in [0,  6  s]#.  What is the average speed of the object?

Amelia Adams · Accepted Answer

#115.3ms^-1#

#E_k = 1/2mv^2#

This is the equation for kinetic energy. It can be rearranged to give #v# as the subject:

#v = sqrt((2E_k)/m)#

We know the mass #m=6# and the energy at the beginning is #E_k = 84kJ = 84000J#, so

#v = sqrt((2*84000)/6) = sqrt(28000) = 167.332ms^-1#

We know that the energy at the end is #12kJ = 12000J#, so

#v = sqrt((2*12000)/6) = sqrt(4000) = 63.25ms^-1#

Now we have the initial and the final velocities.

Work out the average (of anything) by adding them all together and dividing by the total number of things. In this case, we only have two velocities, so:

#bar v = (167.332 + 63.25)/2#

#= 115.291ms^-1#

Layla Ahmed · Answer

undefined To find the average speed of the object, we need to first find the initial and final velocities using the given kinetic energy values. Then, we can use the formula for average speed, which is total distance traveled divided by the total time taken.

Given that kinetic energy $ KE = \frac{1}{2}mv^2 $, we can find the initial and final velocities using the given kinetic energy values.

Initial kinetic energy $ KE_1 = 84 \, kJ $
Final kinetic energy $ KE_2 = 12 \, kJ $

Using the formula for kinetic energy, we can solve for the initial and final velocities:

$ KE_1 = \frac{1}{2}mv_1^2 $  
$ 84 \, kJ = \frac{1}{2} \times 6 \, kg \times v_1^2 $  
$ v_1^2 = \frac{2 \times 84 \, kJ}{6 \, kg} $  
$ v_1^2 = 28 \, m^2/s^2 $  
$ v_1 = \sqrt{28} \, m/s $  
$ v_1 = 5.29 \, m/s $ (approx)

Similarly,  
$ v_2 = \sqrt{\frac{2 \times 12 \, kJ}{6 \, kg}} $  
$ v_2 = \sqrt{4} \, m/s $  
$ v_2 = 2 \, m/s $

Average speed $ v_{avg} = \frac{d}{t} $  
$ v_{avg} = \frac{v_1 + v_2}{2} $  
$ v_{avg} = \frac{5.29 \, m/s + 2 \, m/s}{2} $  
$ v_{avg} = \frac{7.29 \, m/s}{2} $  
$ v_{avg} = 3.65 \, m/s $ (approx)

Therefore, the average speed of the object is approximately $ 3.65 \, m/s $.

Julia Aguilar · Answer

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

Average speed = Total distance traveled / Total time taken

Since the kinetic energy is changing uniformly, we can use the work-energy principle to find the distance traveled. The change in kinetic energy is equal to the work done, which is also equal to the force applied times the distance traveled. Since we have no information about the force, we'll use the formula for kinetic energy:

Kinetic energy = (1/2) * mass * (velocity)^2

From the given kinetic energies, we can find the initial and final velocities using the kinetic energy formula. Then, we can find the average speed by dividing the total distance traveled by the total time taken.

Initial kinetic energy = 84 kJ
Final kinetic energy = 12 kJ
Mass = 6 kg

Initial kinetic energy = (1/2) * 6 * (initial velocity)^2
Final kinetic energy = (1/2) * 6 * (final velocity)^2

Solve for initial velocity and final velocity:

84 kJ = (1/2) * 6 * (initial velocity)^2
12 kJ = (1/2) * 6 * (final velocity)^2

Solve for initial and final velocities:

Initial velocity ≈ 8.164 m/s
Final velocity ≈ 4.899 m/s

Now, we can find the total distance traveled:

Distance traveled = (initial velocity + final velocity) * time

Distance traveled = (8.164 m/s + 4.899 m/s) * 6 s
Distance traveled ≈ 77.796 meters

Now, we can find the average speed:

Average speed = 77.796 meters / 6 s
Average speed ≈ 12.966 m/s