An object has a mass of 4 kg. The object's kinetic energy uniformly changes from 128 KJ to 48 KJ over t in [0, 5 s]. What is the average speed of the object?

1 Answer
Narad T.
Jun 12, 2017

The average speed is =207.9ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =4kg

The initial velocity is =u_1

The initial kinetic energy is 1/2m u_1^2=128000J

The final velocity is =u_2

The final kinetic energy is 1/2m u_2^2=48000J

Therefore,

u_1^2=2/4*128000=64000m^2s^-2

and,

u_2^2=2/4*48000=24000m^2s^-2

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

The points are (0,64000) and (5,24000)

The equation of the line is

v^2-64000=(24000-64000)/5t

v^2=-8000t+64000

So,

v=sqrt((-8000t+64000)

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

(5-0)bar v=int_0^5sqrt((-8000t+64000))dt

5 barv=[((-8000t+64000)^(3/2)/(3/2*8000)]_0^5

=((-8000*5+64000)^(3/2)/(12000))-((-8000*0+64000)^(3/2)/(12000))

=64000^(3/2)/12000-24000^(3/2)/12000

=1039.4

So,

barv=1039.4/5=207.9ms^-1

The average speed is =207.9ms^-1

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