An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 18 KJ to 4KJ over t in [0, 9 s]. What is the average speed of the object?

1 Answer
Narad T.
Mar 16, 2017

The average speed is =59.38ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The initial velocity is =u_1

1/2m u_1^2=18000J

The final velocity is =u_2

1/2m u_2^2=4000J

Therefore,

u_1^2=2/6*18000=6000m^2s^-2

and,

u_2^2=2/6*4000=1333.3m^2s^-2

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

The points are (0,6000) and (9,1333.3)

The equation of the line is

v^2-6000=(1333.3-6000)/9t

v^2=-518.5t+6000

So,

v=sqrt((-518.5t+6000)

We need to calculate the average value of v over t in [0,9]

(9-0)bar v=int_0^9sqrt(-518.5t+6000) dt

9 barv=[(-518.5t+6000)^(3/2)/(3/2*-518.5)]_0^9

=((-518.5*9+6000)^(3/2)/(-777.8))-((-518.5*0+6000)^(3/2)/(-777.8))

=1333.5^(3/2)/-777.8-6000^(3/2)/-777.8

=1/778.5(6000^(3/2)-1333.5^(3/2))

=534.4

So,

barv=534.4/9=59.38ms^-1

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