An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 480 KJ to 108 KJ over t in [0, 8 s]. What is the average speed of the object?

1 Answer
Narad T.
Jun 7, 2017

The average speed is =307.4ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The mass is =6kg

The initial velocity is =u_1

The initial kinetic energy is 1/2m u_1^2=480000J

The final velocity is =u_2

The final kinetic energy is 1/2m u_2^2=108000J

Therefore,

u_1^2=2/6*480000=160000m^2s^-2

and,

u_2^2=2/6*108000=36000m^2s^-2

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

The points are (0,160000) and (8,36000)

The equation of the line is

v^2-160000=(36000-160000)/8t

v^2=-15500t+160000

So,

v=sqrt((-15500t+160000)

We need to calculate the average value of v over t in [0,8]

(8-0)bar v=int_0^8sqrt((-15500t+160000))dt

8 barv=[((-15500t+160000)^(3/2)/(-3/2*15500)]_0^8

=((-15500*8+160000)^(3/2)/(-23250))-((-15500*0+160000)^(3/2)/(-23250))

=160000^(3/2)/23250-36000^(3/2)/23250

=2458.9

So,

barv=2458.9/8=307.4ms^-1

The average speed is =307.4ms^-1

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