An object has a mass of 9 kg. The object's kinetic energy uniformly changes from 135 KJ to 36KJ over t in [0, 4 s]. What is the average speed of the object?

1 Answer
Narad T.
Mar 14, 2017

The average speed is =135.8ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

The initial velocity is =u_1

1/2m u_1^2=135000J

The final velocity is =u_2

1/2m u_2^2=36000J

Therefore,

u_1^2=2/9*135000=30000m^2s^-2

and,

u_2^2=2/9*36000=8000m^2s^-2

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

The points are (0,30000) and (4,8000)

The equation of the line is

v^2-30000=(8000-30000)/4t

v^2=-5500t+30000

So,

v=sqrt((-5500t+30000)

We need to calculate the average value of v over t in [0,4]

(4-0)bar v=int_0^5sqrt(-5500t+30000) dt

4 barv=[(-5500t+30000)^(3/2)/(3/2*-5500)]_0^4

=((-5500*4+30000)^(3/2)/(-8250))-((-5500*0+30000)^(3/2)/(-8250))

=8000^(3/2)/-8250-30000^(3/2)/-8250

=1/8250(30000^(3/2)-8000^(3/2)))

=543.1

So,

barv=543.1/4=135.8ms^-1

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