An object has a mass of 6 kg. The object's kinetic energy uniformly changes from 540 KJ to 36 KJ over t in [0, 4 s]. What is the average speed of the object?

1 Answer
Narad T.
Apr 10, 2017

The average speed is =297.8ms^-1

Explanation:

The kinetic energy is

KE=1/2mv^2

mass is =6kg

The initial velocity is =u_1

1/2m u_1^2=540000J

The final velocity is =u_2

1/2m u_2^2=36000J

Therefore,

u_1^2=2/6*540000=180000m^2s^-2

and,

u_2^2=2/6*36000=12000m^2s^-2

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

The points are (0,180000) and (4,12000)

The equation of the line is

v^2-180000=(12000-180000)/4t

v^2=-42000t+180000

So,

v=sqrt((-42000t+180000)

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

(4-0)bar v=int_0^4sqrt((-42000t+180000))dt

4 barv=[((-42000t+180000)^(3/2)/(-3/2*42000)]_0^4

=((-42000*4+180000)^(3/2)/(-63000))-((-42000*0+180000)^(3/2)/(-63000))

=180000^(3/2)/63000-12000^(3/2)/63000

=1191.3

So,

barv=1191.3/4=297.8ms^-1

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