An object has a mass of 8 kg. The object's kinetic energy uniformly changes from 24 KJ to 42KJ over t in [0, 3 s]. What is the average speed of the object?

1 Answer
Narad T.
Jun 14, 2017

The average speed is =90.54ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =8kg

The initial velocity is =u_1

The initial kinetic energy is 1/2m u_1^2=24000J

The final velocity is =u_2

The final kinetic energy is 1/2m u_2^2=42000J

Therefore,

u_1^2=2/8*24000=6000m^2s^-2

and,

u_2^2=2/8*42000=10500m^2s^-2

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

The points are (0,6000) and (3,10500)

The equation of the line is

v^2-6000=(10500-6000)/3t

v^2=1500t+6000

So,

v=sqrt((1500t+6000)

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

(3-0)bar v=int_0^3sqrt((1500t+6000))dt

3 barv=[((1500t+6000)^(3/2)/(3/2*1500)]_0^3

=((1500*3+6000)^(3/2)/(2250))-((1500*0+6000)^(3/2)/(2250))

=10500^(3/2)/2250-6000^(3/2)/2250

=271.63

So,

barv=271.63/3=90.54ms^-1

The average speed is =90.54ms^-1

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